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Q&A Remove entries by two-column ID everywhere, that meet a condition somewhere

Answer After re-encountering this problem and searching for a solution with a less generic "pandas check if column pair is found in other df" (forgetting I had previously encountered this proble...

posted 1y ago by mcp‭  ·  edited 1y ago by mcp‭

Answer
#3: Post edited by user avatar mcp‭ · 2023-07-05T16:33:42Z (over 1 year ago)
Remove duplicates in all methods for consistency and update timings
  • # Answer
  • After re-encountering this problem and searching for a solution with a
  • less generic "pandas check if column pair is found in other df"
  • (forgetting I had previously encountered this problem altogether), I
  • found some inspiration that led me to a solid solution.
  • It is fastest, and most correct to form a column containing tuple values
  • of the value pair in question, and use `isin()` to find the matches.
  • This column can be later dropped after use.
  • ```py
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • ```
  • ```txt
  • Die Cell Current Resistance Die, Cell
  • 0 1.0 0.0 100.0 104155.0 (1, 0)
  • 1 1.0 0.0 200.0 2542.0 (1, 0)
  • 2 1.0 0.0 300.0 206302.0 (1, 0)
  • 3 NaN NaN NaN NaN NaN
  • 4 NaN NaN NaN NaN NaN
  • 5 NaN NaN NaN NaN NaN
  • 6 1.0 2.0 100.0 141502.0 (1, 2)
  • 7 1.0 2.0 200.0 40422.0 (1, 2)
  • 8 1.0 2.0 300.0 31066.0 (1, 2)
  • 9 1.0 3.0 100.0 108234.0 (1, 3)
  • 10 1.0 3.0 200.0 140520.0 (1, 3)
  • 11 1.0 3.0 300.0 43586.0 (1, 3)
  • 12 1.0 4.0 100.0 138305.0 (1, 4)
  • 13 1.0 4.0 200.0 88725.0 (1, 4)
  • 14 1.0 4.0 300.0 239517.0 (1, 4)
  • 15 NaN NaN NaN NaN NaN
  • 16 NaN NaN NaN NaN NaN
  • 17 NaN NaN NaN NaN NaN
  • 18 1.0 6.0 100.0 125984.0 (1, 6)
  • 19 1.0 6.0 200.0 37036.0 (1, 6)
  • 20 1.0 6.0 300.0 179742.0 (1, 6)
  • 21 NaN NaN NaN NaN NaN
  • 22 NaN NaN NaN NaN NaN
  • 23 NaN NaN NaN NaN NaN
  • 24 NaN NaN NaN NaN NaN
  • 25 NaN NaN NaN NaN NaN
  • 26 NaN NaN NaN NaN NaN
  • 27 1.0 9.0 100.0 186133.0 (1, 9)
  • 28 1.0 9.0 200.0 863.0 (1, 9)
  • 29 1.0 9.0 300.0 235060.0 (1, 9)
  • 30 2.0 0.0 100.0 217692.0 (2, 0)
  • 31 2.0 0.0 200.0 192709.0 (2, 0)
  • 32 2.0 0.0 300.0 44718.0 (2, 0)
  • 33 NaN NaN NaN NaN NaN
  • 34 NaN NaN NaN NaN NaN
  • 35 NaN NaN NaN NaN NaN
  • 36 2.0 2.0 100.0 144522.0 (2, 2)
  • 37 2.0 2.0 200.0 184146.0 (2, 2)
  • 38 2.0 2.0 300.0 58155.0 (2, 2)
  • 39 2.0 3.0 100.0 130899.0 (2, 3)
  • 40 2.0 3.0 200.0 177347.0 (2, 3)
  • 41 2.0 3.0 300.0 206209.0 (2, 3)
  • 42 2.0 4.0 100.0 201781.0 (2, 4)
  • 43 2.0 4.0 200.0 58077.0 (2, 4)
  • 44 2.0 4.0 300.0 218298.0 (2, 4)
  • 45 NaN NaN NaN NaN NaN
  • 46 NaN NaN NaN NaN NaN
  • 47 NaN NaN NaN NaN NaN
  • 48 NaN NaN NaN NaN NaN
  • 49 NaN NaN NaN NaN NaN
  • 50 NaN NaN NaN NaN NaN
  • 51 2.0 7.0 100.0 239666.0 (2, 7)
  • 52 2.0 7.0 200.0 10384.0 (2, 7)
  • 53 2.0 7.0 300.0 41034.0 (2, 7)
  • 54 2.0 8.0 100.0 245824.0 (2, 8)
  • 55 2.0 8.0 200.0 208052.0 (2, 8)
  • 56 2.0 8.0 300.0 37568.0 (2, 8)
  • 57 NaN NaN NaN NaN NaN
  • 58 NaN NaN NaN NaN NaN
  • 59 NaN NaN NaN NaN NaN
  • ```
  • # Timings
  • I determined it to be fastest by testing the following methods:
  • ```py
  • import time
  • import random
  • import numpy as np
  • import pandas as pd
  • from itertools import product
  • random.seed(12345)
  • # dies = [1, 2]
  • # cells = list(range(10))
  • # currents = [100, 200, 300]
  • dies = list(range(10))
  • cells = list(range(1000))
  • currents = list(range(100, 800 + 100, 100))
  • dcc = list(product(dies, cells, currents))
  • resistances = random.choices(range(250000 + 1), k=len(dcc))
  • df = pd.DataFrame(dcc, columns=["Die", "Cell", "Current"])
  • df["Resistance"] = resistances
  • b100 = df[(df["Current"] == 100) & (df["Resistance"] < 100000)]
  • print(f"df:\n{df}\n")
  • print(f"b100:\n{b100}\n")
  • # --
  • df_og = df.copy()
  • b100_og = b100.copy()
  • # --
  • start = time.perf_counter()
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • for die, cell in zip(b100["Die"], b100["Cell"]):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_zip_time = stop - start
  • # --
  • # INCORRECT
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df = df.set_index(["Die", "Cell", "Current"])
  • df.loc[(b100["Die"], b100["Cell"], slice(None)), :] = np.nan
  • stop = time.perf_counter()
  • slice_time = stop - start
  • # INCORRECT
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • stop = time.perf_counter()
  • tuple_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(
  • df[["Die", "Cell"]].apply(tuple, axis="columns").isin(
  • b100[["Die", "Cell"]].apply(tuple, axis="columns")))
  • stop = time.perf_counter()
  • tuple_fly_time = stop - start
  • # --
  • print(f"Loop time: {loop_time:.2f} seconds")
  • print(f"Loop no dup time: {loop_no_dup_time:.2f} seconds")
  • print(f"Loop no dup zip time: {loop_no_dup_zip_time:.2f} seconds")
  • print(f"Slice time: {slice_time:.2f} seconds")
  • print(f"Tuple time: {tuple_time:.2f} seconds")
  • print(f"Tuple fly time: {tuple_fly_time:.2f} seconds")
  • ```
  • ```txt
  • Loop time: 14.78 seconds
  • Loop no dup time: 14.60 seconds
  • Loop no dup zip time: 15.54 seconds
  • Slice time: 0.16 seconds
  • Tuple time: 0.04 seconds
  • Tuple fly time: 0.31 seconds
  • Intel(R) Core(TM) i7-10510U CPU @ 1.80GHz
  • ```
  • ## Notes
  • Slicing is a close second on speed, but produces an incorrect result.
  • Same behavior as (mentioned in [Tried](https://software.codidact.com/posts/288068)):
  • ```py
  • df = df.mask((df["Die"].isin(b100["Die"])) & (df["Cell"].isin(b100["Cell"])))
  • ```
  • The use of `drop_duplicates()` is technically not needed, since `b100`
  • is created at a single current, with no duplicate die, cell pairs.
  • Despite this, it is left in place for my own convenience.
  • ---
  • Inspired by:
  • https://stackoverflow.com/a/16068497
  • https://stackoverflow.com/a/70355304
  • # Answer
  • After re-encountering this problem and searching for a solution with a
  • less generic "pandas check if column pair is found in other df"
  • (forgetting I had previously encountered this problem altogether), I
  • found some inspiration that led me to a solid solution.
  • It is fastest, and most correct to form a column containing tuple values
  • of the value pair in question, and use `isin()` to find the matches.
  • This column can be later dropped after use.
  • ```py
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • ```
  • ```txt
  • Die Cell Current Resistance Die, Cell
  • 0 1.0 0.0 100.0 104155.0 (1, 0)
  • 1 1.0 0.0 200.0 2542.0 (1, 0)
  • 2 1.0 0.0 300.0 206302.0 (1, 0)
  • 3 NaN NaN NaN NaN NaN
  • 4 NaN NaN NaN NaN NaN
  • 5 NaN NaN NaN NaN NaN
  • 6 1.0 2.0 100.0 141502.0 (1, 2)
  • 7 1.0 2.0 200.0 40422.0 (1, 2)
  • 8 1.0 2.0 300.0 31066.0 (1, 2)
  • 9 1.0 3.0 100.0 108234.0 (1, 3)
  • 10 1.0 3.0 200.0 140520.0 (1, 3)
  • 11 1.0 3.0 300.0 43586.0 (1, 3)
  • 12 1.0 4.0 100.0 138305.0 (1, 4)
  • 13 1.0 4.0 200.0 88725.0 (1, 4)
  • 14 1.0 4.0 300.0 239517.0 (1, 4)
  • 15 NaN NaN NaN NaN NaN
  • 16 NaN NaN NaN NaN NaN
  • 17 NaN NaN NaN NaN NaN
  • 18 1.0 6.0 100.0 125984.0 (1, 6)
  • 19 1.0 6.0 200.0 37036.0 (1, 6)
  • 20 1.0 6.0 300.0 179742.0 (1, 6)
  • 21 NaN NaN NaN NaN NaN
  • 22 NaN NaN NaN NaN NaN
  • 23 NaN NaN NaN NaN NaN
  • 24 NaN NaN NaN NaN NaN
  • 25 NaN NaN NaN NaN NaN
  • 26 NaN NaN NaN NaN NaN
  • 27 1.0 9.0 100.0 186133.0 (1, 9)
  • 28 1.0 9.0 200.0 863.0 (1, 9)
  • 29 1.0 9.0 300.0 235060.0 (1, 9)
  • 30 2.0 0.0 100.0 217692.0 (2, 0)
  • 31 2.0 0.0 200.0 192709.0 (2, 0)
  • 32 2.0 0.0 300.0 44718.0 (2, 0)
  • 33 NaN NaN NaN NaN NaN
  • 34 NaN NaN NaN NaN NaN
  • 35 NaN NaN NaN NaN NaN
  • 36 2.0 2.0 100.0 144522.0 (2, 2)
  • 37 2.0 2.0 200.0 184146.0 (2, 2)
  • 38 2.0 2.0 300.0 58155.0 (2, 2)
  • 39 2.0 3.0 100.0 130899.0 (2, 3)
  • 40 2.0 3.0 200.0 177347.0 (2, 3)
  • 41 2.0 3.0 300.0 206209.0 (2, 3)
  • 42 2.0 4.0 100.0 201781.0 (2, 4)
  • 43 2.0 4.0 200.0 58077.0 (2, 4)
  • 44 2.0 4.0 300.0 218298.0 (2, 4)
  • 45 NaN NaN NaN NaN NaN
  • 46 NaN NaN NaN NaN NaN
  • 47 NaN NaN NaN NaN NaN
  • 48 NaN NaN NaN NaN NaN
  • 49 NaN NaN NaN NaN NaN
  • 50 NaN NaN NaN NaN NaN
  • 51 2.0 7.0 100.0 239666.0 (2, 7)
  • 52 2.0 7.0 200.0 10384.0 (2, 7)
  • 53 2.0 7.0 300.0 41034.0 (2, 7)
  • 54 2.0 8.0 100.0 245824.0 (2, 8)
  • 55 2.0 8.0 200.0 208052.0 (2, 8)
  • 56 2.0 8.0 300.0 37568.0 (2, 8)
  • 57 NaN NaN NaN NaN NaN
  • 58 NaN NaN NaN NaN NaN
  • 59 NaN NaN NaN NaN NaN
  • ```
  • # Timings
  • I determined it to be fastest by testing the following methods:
  • ```py
  • import time
  • import random
  • import numpy as np
  • import pandas as pd
  • from itertools import product
  • random.seed(12345)
  • # dies = [1, 2]
  • # cells = list(range(10))
  • # currents = [100, 200, 300]
  • dies = list(range(10))
  • cells = list(range(1000))
  • currents = list(range(100, 800 + 100, 100))
  • dcc = list(product(dies, cells, currents))
  • resistances = random.choices(range(250000 + 1), k=len(dcc))
  • df = pd.DataFrame(dcc, columns=["Die", "Cell", "Current"])
  • df["Resistance"] = resistances
  • b100 = df[(df["Current"] == 100) & (df["Resistance"] < 100000)]
  • print(f"df:\n{df}\n")
  • print(f"b100:\n{b100}\n")
  • # --
  • df_og = df.copy()
  • b100_og = b100.copy()
  • # --
  • start = time.perf_counter()
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • for die, cell in zip(b100["Die"], b100["Cell"]):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_zip_time = stop - start
  • # --
  • # INCORRECT
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df = df.set_index(["Die", "Cell", "Current"])
  • df.loc[(b100["Die"], b100["Cell"], slice(None)), :] = np.nan
  • stop = time.perf_counter()
  • slice_time = stop - start
  • # INCORRECT
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • stop = time.perf_counter()
  • tuple_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(
  • df[["Die", "Cell"]].apply(tuple, axis="columns").isin(
  • b100[["Die", "Cell"]].apply(tuple, axis="columns")))
  • stop = time.perf_counter()
  • tuple_fly_time = stop - start
  • # --
  • print(f"Loop time: {loop_time:.2f} seconds")
  • print(f"Loop no dup time: {loop_no_dup_time:.2f} seconds")
  • print(f"Loop no dup zip time: {loop_no_dup_zip_time:.2f} seconds")
  • print(f"Slice time: {slice_time:.2f} seconds")
  • print(f"Tuple time: {tuple_time:.2f} seconds")
  • print(f"Tuple fly time: {tuple_fly_time:.2f} seconds")
  • ```
  • ```txt
  • Loop time: 14.01 seconds
  • Loop no dup time: 14.66 seconds
  • Loop no dup zip time: 16.66 seconds
  • Slice time: 0.16 seconds
  • Tuple time: 0.03 seconds
  • Tuple fly time: 0.28 seconds
  • Intel(R) Core(TM) i7-10510U CPU @ 1.80GHz
  • ```
  • ## Notes
  • Slicing is a close second on speed, but produces an incorrect result.
  • Same behavior as (mentioned in [Tried](https://software.codidact.com/posts/288068)):
  • ```py
  • df = df.mask((df["Die"].isin(b100["Die"])) & (df["Cell"].isin(b100["Cell"])))
  • ```
  • The use of `drop_duplicates()` is technically not needed, since `b100`
  • is created at a single current, with no duplicate die, cell pairs.
  • Despite this, it is left in place for my own convenience.
  • ---
  • Inspired by:
  • https://stackoverflow.com/a/16068497
  • https://stackoverflow.com/a/70355304
#2: Post edited by user avatar mcp‭ · 2023-07-05T15:57:41Z (over 1 year ago)
Fix grammar and reference Tried
  • # Answer
  • After re-encountering this problem and searching for a solution with a
  • less generic "pandas check if column pair is found in other df"
  • (forgetting I had previously encountered this problem altogether), I
  • found some inspiration that led me to a solid solution.
  • It is fastest, and most correct to form a column containing tuple values
  • of the value pair in question, and using `isin()` to find the matches.
  • This column can be later dropped after use.
  • ```py
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • ```
  • ```txt
  • Die Cell Current Resistance Die, Cell
  • 0 1.0 0.0 100.0 104155.0 (1, 0)
  • 1 1.0 0.0 200.0 2542.0 (1, 0)
  • 2 1.0 0.0 300.0 206302.0 (1, 0)
  • 3 NaN NaN NaN NaN NaN
  • 4 NaN NaN NaN NaN NaN
  • 5 NaN NaN NaN NaN NaN
  • 6 1.0 2.0 100.0 141502.0 (1, 2)
  • 7 1.0 2.0 200.0 40422.0 (1, 2)
  • 8 1.0 2.0 300.0 31066.0 (1, 2)
  • 9 1.0 3.0 100.0 108234.0 (1, 3)
  • 10 1.0 3.0 200.0 140520.0 (1, 3)
  • 11 1.0 3.0 300.0 43586.0 (1, 3)
  • 12 1.0 4.0 100.0 138305.0 (1, 4)
  • 13 1.0 4.0 200.0 88725.0 (1, 4)
  • 14 1.0 4.0 300.0 239517.0 (1, 4)
  • 15 NaN NaN NaN NaN NaN
  • 16 NaN NaN NaN NaN NaN
  • 17 NaN NaN NaN NaN NaN
  • 18 1.0 6.0 100.0 125984.0 (1, 6)
  • 19 1.0 6.0 200.0 37036.0 (1, 6)
  • 20 1.0 6.0 300.0 179742.0 (1, 6)
  • 21 NaN NaN NaN NaN NaN
  • 22 NaN NaN NaN NaN NaN
  • 23 NaN NaN NaN NaN NaN
  • 24 NaN NaN NaN NaN NaN
  • 25 NaN NaN NaN NaN NaN
  • 26 NaN NaN NaN NaN NaN
  • 27 1.0 9.0 100.0 186133.0 (1, 9)
  • 28 1.0 9.0 200.0 863.0 (1, 9)
  • 29 1.0 9.0 300.0 235060.0 (1, 9)
  • 30 2.0 0.0 100.0 217692.0 (2, 0)
  • 31 2.0 0.0 200.0 192709.0 (2, 0)
  • 32 2.0 0.0 300.0 44718.0 (2, 0)
  • 33 NaN NaN NaN NaN NaN
  • 34 NaN NaN NaN NaN NaN
  • 35 NaN NaN NaN NaN NaN
  • 36 2.0 2.0 100.0 144522.0 (2, 2)
  • 37 2.0 2.0 200.0 184146.0 (2, 2)
  • 38 2.0 2.0 300.0 58155.0 (2, 2)
  • 39 2.0 3.0 100.0 130899.0 (2, 3)
  • 40 2.0 3.0 200.0 177347.0 (2, 3)
  • 41 2.0 3.0 300.0 206209.0 (2, 3)
  • 42 2.0 4.0 100.0 201781.0 (2, 4)
  • 43 2.0 4.0 200.0 58077.0 (2, 4)
  • 44 2.0 4.0 300.0 218298.0 (2, 4)
  • 45 NaN NaN NaN NaN NaN
  • 46 NaN NaN NaN NaN NaN
  • 47 NaN NaN NaN NaN NaN
  • 48 NaN NaN NaN NaN NaN
  • 49 NaN NaN NaN NaN NaN
  • 50 NaN NaN NaN NaN NaN
  • 51 2.0 7.0 100.0 239666.0 (2, 7)
  • 52 2.0 7.0 200.0 10384.0 (2, 7)
  • 53 2.0 7.0 300.0 41034.0 (2, 7)
  • 54 2.0 8.0 100.0 245824.0 (2, 8)
  • 55 2.0 8.0 200.0 208052.0 (2, 8)
  • 56 2.0 8.0 300.0 37568.0 (2, 8)
  • 57 NaN NaN NaN NaN NaN
  • 58 NaN NaN NaN NaN NaN
  • 59 NaN NaN NaN NaN NaN
  • ```
  • # Timings
  • I determined it to be fastest by testing the following methods:
  • ```py
  • import time
  • import random
  • import numpy as np
  • import pandas as pd
  • from itertools import product
  • random.seed(12345)
  • # dies = [1, 2]
  • # cells = list(range(10))
  • # currents = [100, 200, 300]
  • dies = list(range(10))
  • cells = list(range(1000))
  • currents = list(range(100, 800 + 100, 100))
  • dcc = list(product(dies, cells, currents))
  • resistances = random.choices(range(250000 + 1), k=len(dcc))
  • df = pd.DataFrame(dcc, columns=["Die", "Cell", "Current"])
  • df["Resistance"] = resistances
  • b100 = df[(df["Current"] == 100) & (df["Resistance"] < 100000)]
  • print(f"df:\n{df}\n")
  • print(f"b100:\n{b100}\n")
  • # --
  • df_og = df.copy()
  • b100_og = b100.copy()
  • # --
  • start = time.perf_counter()
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • for die, cell in zip(b100["Die"], b100["Cell"]):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_zip_time = stop - start
  • # --
  • # INCORRECT
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df = df.set_index(["Die", "Cell", "Current"])
  • df.loc[(b100["Die"], b100["Cell"], slice(None)), :] = np.nan
  • stop = time.perf_counter()
  • slice_time = stop - start
  • # INCORRECT
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • stop = time.perf_counter()
  • tuple_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(
  • df[["Die", "Cell"]].apply(tuple, axis="columns").isin(
  • b100[["Die", "Cell"]].apply(tuple, axis="columns")))
  • stop = time.perf_counter()
  • tuple_fly_time = stop - start
  • # --
  • print(f"Loop time: {loop_time:.2f} seconds")
  • print(f"Loop no dup time: {loop_no_dup_time:.2f} seconds")
  • print(f"Loop no dup zip time: {loop_no_dup_zip_time:.2f} seconds")
  • print(f"Slice time: {slice_time:.2f} seconds")
  • print(f"Tuple time: {tuple_time:.2f} seconds")
  • print(f"Tuple fly time: {tuple_fly_time:.2f} seconds")
  • ```
  • ```txt
  • Loop time: 14.78 seconds
  • Loop no dup time: 14.60 seconds
  • Loop no dup zip time: 15.54 seconds
  • Slice time: 0.16 seconds
  • Tuple time: 0.04 seconds
  • Tuple fly time: 0.31 seconds
  • Intel(R) Core(TM) i7-10510U CPU @ 1.80GHz
  • ```
  • ## Notes
  • Slicing is a close second on speed, but produces an incorrect result.
  • Same behavior as:
  • ```py
  • df = df.mask((df["Die"].isin(b100["Die"])) & (df["Cell"].isin(b100["Cell"])))
  • ```
  • The use of `drop_duplicates()` is technically not needed, since `b100`
  • is created at a single current, with no duplicate die, cell pairs.
  • Despite this, it is left in place for my own convenience.
  • ---
  • Inspired by:
  • https://stackoverflow.com/a/16068497
  • https://stackoverflow.com/a/70355304
  • # Answer
  • After re-encountering this problem and searching for a solution with a
  • less generic "pandas check if column pair is found in other df"
  • (forgetting I had previously encountered this problem altogether), I
  • found some inspiration that led me to a solid solution.
  • It is fastest, and most correct to form a column containing tuple values
  • of the value pair in question, and use `isin()` to find the matches.
  • This column can be later dropped after use.
  • ```py
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • ```
  • ```txt
  • Die Cell Current Resistance Die, Cell
  • 0 1.0 0.0 100.0 104155.0 (1, 0)
  • 1 1.0 0.0 200.0 2542.0 (1, 0)
  • 2 1.0 0.0 300.0 206302.0 (1, 0)
  • 3 NaN NaN NaN NaN NaN
  • 4 NaN NaN NaN NaN NaN
  • 5 NaN NaN NaN NaN NaN
  • 6 1.0 2.0 100.0 141502.0 (1, 2)
  • 7 1.0 2.0 200.0 40422.0 (1, 2)
  • 8 1.0 2.0 300.0 31066.0 (1, 2)
  • 9 1.0 3.0 100.0 108234.0 (1, 3)
  • 10 1.0 3.0 200.0 140520.0 (1, 3)
  • 11 1.0 3.0 300.0 43586.0 (1, 3)
  • 12 1.0 4.0 100.0 138305.0 (1, 4)
  • 13 1.0 4.0 200.0 88725.0 (1, 4)
  • 14 1.0 4.0 300.0 239517.0 (1, 4)
  • 15 NaN NaN NaN NaN NaN
  • 16 NaN NaN NaN NaN NaN
  • 17 NaN NaN NaN NaN NaN
  • 18 1.0 6.0 100.0 125984.0 (1, 6)
  • 19 1.0 6.0 200.0 37036.0 (1, 6)
  • 20 1.0 6.0 300.0 179742.0 (1, 6)
  • 21 NaN NaN NaN NaN NaN
  • 22 NaN NaN NaN NaN NaN
  • 23 NaN NaN NaN NaN NaN
  • 24 NaN NaN NaN NaN NaN
  • 25 NaN NaN NaN NaN NaN
  • 26 NaN NaN NaN NaN NaN
  • 27 1.0 9.0 100.0 186133.0 (1, 9)
  • 28 1.0 9.0 200.0 863.0 (1, 9)
  • 29 1.0 9.0 300.0 235060.0 (1, 9)
  • 30 2.0 0.0 100.0 217692.0 (2, 0)
  • 31 2.0 0.0 200.0 192709.0 (2, 0)
  • 32 2.0 0.0 300.0 44718.0 (2, 0)
  • 33 NaN NaN NaN NaN NaN
  • 34 NaN NaN NaN NaN NaN
  • 35 NaN NaN NaN NaN NaN
  • 36 2.0 2.0 100.0 144522.0 (2, 2)
  • 37 2.0 2.0 200.0 184146.0 (2, 2)
  • 38 2.0 2.0 300.0 58155.0 (2, 2)
  • 39 2.0 3.0 100.0 130899.0 (2, 3)
  • 40 2.0 3.0 200.0 177347.0 (2, 3)
  • 41 2.0 3.0 300.0 206209.0 (2, 3)
  • 42 2.0 4.0 100.0 201781.0 (2, 4)
  • 43 2.0 4.0 200.0 58077.0 (2, 4)
  • 44 2.0 4.0 300.0 218298.0 (2, 4)
  • 45 NaN NaN NaN NaN NaN
  • 46 NaN NaN NaN NaN NaN
  • 47 NaN NaN NaN NaN NaN
  • 48 NaN NaN NaN NaN NaN
  • 49 NaN NaN NaN NaN NaN
  • 50 NaN NaN NaN NaN NaN
  • 51 2.0 7.0 100.0 239666.0 (2, 7)
  • 52 2.0 7.0 200.0 10384.0 (2, 7)
  • 53 2.0 7.0 300.0 41034.0 (2, 7)
  • 54 2.0 8.0 100.0 245824.0 (2, 8)
  • 55 2.0 8.0 200.0 208052.0 (2, 8)
  • 56 2.0 8.0 300.0 37568.0 (2, 8)
  • 57 NaN NaN NaN NaN NaN
  • 58 NaN NaN NaN NaN NaN
  • 59 NaN NaN NaN NaN NaN
  • ```
  • # Timings
  • I determined it to be fastest by testing the following methods:
  • ```py
  • import time
  • import random
  • import numpy as np
  • import pandas as pd
  • from itertools import product
  • random.seed(12345)
  • # dies = [1, 2]
  • # cells = list(range(10))
  • # currents = [100, 200, 300]
  • dies = list(range(10))
  • cells = list(range(1000))
  • currents = list(range(100, 800 + 100, 100))
  • dcc = list(product(dies, cells, currents))
  • resistances = random.choices(range(250000 + 1), k=len(dcc))
  • df = pd.DataFrame(dcc, columns=["Die", "Cell", "Current"])
  • df["Resistance"] = resistances
  • b100 = df[(df["Current"] == 100) & (df["Resistance"] < 100000)]
  • print(f"df:\n{df}\n")
  • print(f"b100:\n{b100}\n")
  • # --
  • df_og = df.copy()
  • b100_og = b100.copy()
  • # --
  • start = time.perf_counter()
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • rm_dies = b100["Die"].to_list()
  • rm_cells = b100["Cell"].to_list()
  • for die, cell in zip(rm_dies, rm_cells):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • for die, cell in zip(b100["Die"], b100["Cell"]):
  • df = df.mask((df["Die"] == die) & (df["Cell"] == cell))
  • stop = time.perf_counter()
  • loop_no_dup_zip_time = stop - start
  • # --
  • # INCORRECT
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • b100 = b100.drop_duplicates(["Die", "Cell"])
  • df = df.set_index(["Die", "Cell", "Current"])
  • df.loc[(b100["Die"], b100["Cell"], slice(None)), :] = np.nan
  • stop = time.perf_counter()
  • slice_time = stop - start
  • # INCORRECT
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
  • stop = time.perf_counter()
  • tuple_time = stop - start
  • # --
  • df = df_og.copy()
  • b100 = b100_og.copy()
  • start = time.perf_counter()
  • df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
  • b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))
  • df = df.mask(
  • df[["Die", "Cell"]].apply(tuple, axis="columns").isin(
  • b100[["Die", "Cell"]].apply(tuple, axis="columns")))
  • stop = time.perf_counter()
  • tuple_fly_time = stop - start
  • # --
  • print(f"Loop time: {loop_time:.2f} seconds")
  • print(f"Loop no dup time: {loop_no_dup_time:.2f} seconds")
  • print(f"Loop no dup zip time: {loop_no_dup_zip_time:.2f} seconds")
  • print(f"Slice time: {slice_time:.2f} seconds")
  • print(f"Tuple time: {tuple_time:.2f} seconds")
  • print(f"Tuple fly time: {tuple_fly_time:.2f} seconds")
  • ```
  • ```txt
  • Loop time: 14.78 seconds
  • Loop no dup time: 14.60 seconds
  • Loop no dup zip time: 15.54 seconds
  • Slice time: 0.16 seconds
  • Tuple time: 0.04 seconds
  • Tuple fly time: 0.31 seconds
  • Intel(R) Core(TM) i7-10510U CPU @ 1.80GHz
  • ```
  • ## Notes
  • Slicing is a close second on speed, but produces an incorrect result.
  • Same behavior as (mentioned in [Tried](https://software.codidact.com/posts/288068)):
  • ```py
  • df = df.mask((df["Die"].isin(b100["Die"])) & (df["Cell"].isin(b100["Cell"])))
  • ```
  • The use of `drop_duplicates()` is technically not needed, since `b100`
  • is created at a single current, with no duplicate die, cell pairs.
  • Despite this, it is left in place for my own convenience.
  • ---
  • Inspired by:
  • https://stackoverflow.com/a/16068497
  • https://stackoverflow.com/a/70355304
#1: Initial revision by user avatar mcp‭ · 2023-07-04T22:54:22Z (over 1 year ago)
# Answer

After re-encountering this problem and searching for a solution with a
less generic "pandas check if column pair is found in other df"
(forgetting I had previously encountered this problem altogether), I
found some inspiration that led me to a solid solution.

It is fastest, and most correct to form a column containing tuple values
of the value pair in question, and using `isin()` to find the matches.
This column can be later dropped after use.

```py
df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))

df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))
```

```txt
    Die  Cell  Current  Resistance Die, Cell
0   1.0   0.0    100.0    104155.0    (1, 0)
1   1.0   0.0    200.0      2542.0    (1, 0)
2   1.0   0.0    300.0    206302.0    (1, 0)
3   NaN   NaN      NaN         NaN       NaN
4   NaN   NaN      NaN         NaN       NaN
5   NaN   NaN      NaN         NaN       NaN
6   1.0   2.0    100.0    141502.0    (1, 2)
7   1.0   2.0    200.0     40422.0    (1, 2)
8   1.0   2.0    300.0     31066.0    (1, 2)
9   1.0   3.0    100.0    108234.0    (1, 3)
10  1.0   3.0    200.0    140520.0    (1, 3)
11  1.0   3.0    300.0     43586.0    (1, 3)
12  1.0   4.0    100.0    138305.0    (1, 4)
13  1.0   4.0    200.0     88725.0    (1, 4)
14  1.0   4.0    300.0    239517.0    (1, 4)
15  NaN   NaN      NaN         NaN       NaN
16  NaN   NaN      NaN         NaN       NaN
17  NaN   NaN      NaN         NaN       NaN
18  1.0   6.0    100.0    125984.0    (1, 6)
19  1.0   6.0    200.0     37036.0    (1, 6)
20  1.0   6.0    300.0    179742.0    (1, 6)
21  NaN   NaN      NaN         NaN       NaN
22  NaN   NaN      NaN         NaN       NaN
23  NaN   NaN      NaN         NaN       NaN
24  NaN   NaN      NaN         NaN       NaN
25  NaN   NaN      NaN         NaN       NaN
26  NaN   NaN      NaN         NaN       NaN
27  1.0   9.0    100.0    186133.0    (1, 9)
28  1.0   9.0    200.0       863.0    (1, 9)
29  1.0   9.0    300.0    235060.0    (1, 9)
30  2.0   0.0    100.0    217692.0    (2, 0)
31  2.0   0.0    200.0    192709.0    (2, 0)
32  2.0   0.0    300.0     44718.0    (2, 0)
33  NaN   NaN      NaN         NaN       NaN
34  NaN   NaN      NaN         NaN       NaN
35  NaN   NaN      NaN         NaN       NaN
36  2.0   2.0    100.0    144522.0    (2, 2)
37  2.0   2.0    200.0    184146.0    (2, 2)
38  2.0   2.0    300.0     58155.0    (2, 2)
39  2.0   3.0    100.0    130899.0    (2, 3)
40  2.0   3.0    200.0    177347.0    (2, 3)
41  2.0   3.0    300.0    206209.0    (2, 3)
42  2.0   4.0    100.0    201781.0    (2, 4)
43  2.0   4.0    200.0     58077.0    (2, 4)
44  2.0   4.0    300.0    218298.0    (2, 4)
45  NaN   NaN      NaN         NaN       NaN
46  NaN   NaN      NaN         NaN       NaN
47  NaN   NaN      NaN         NaN       NaN
48  NaN   NaN      NaN         NaN       NaN
49  NaN   NaN      NaN         NaN       NaN
50  NaN   NaN      NaN         NaN       NaN
51  2.0   7.0    100.0    239666.0    (2, 7)
52  2.0   7.0    200.0     10384.0    (2, 7)
53  2.0   7.0    300.0     41034.0    (2, 7)
54  2.0   8.0    100.0    245824.0    (2, 8)
55  2.0   8.0    200.0    208052.0    (2, 8)
56  2.0   8.0    300.0     37568.0    (2, 8)
57  NaN   NaN      NaN         NaN       NaN
58  NaN   NaN      NaN         NaN       NaN
59  NaN   NaN      NaN         NaN       NaN
```

# Timings

I determined it to be fastest by testing the following methods:
```py
import time
import random
import numpy as np
import pandas as pd

from itertools import product


random.seed(12345)

# dies = [1, 2]
# cells = list(range(10))
# currents = [100, 200, 300]

dies = list(range(10))
cells = list(range(1000))
currents = list(range(100, 800 + 100, 100))

dcc = list(product(dies, cells, currents))
resistances = random.choices(range(250000 + 1), k=len(dcc))

df = pd.DataFrame(dcc, columns=["Die", "Cell", "Current"])
df["Resistance"] = resistances

b100 = df[(df["Current"] == 100) & (df["Resistance"] < 100000)]

print(f"df:\n{df}\n")
print(f"b100:\n{b100}\n")

# --

df_og = df.copy()
b100_og = b100.copy()

# --

start = time.perf_counter()

rm_dies = b100["Die"].to_list()
rm_cells = b100["Cell"].to_list()

for die, cell in zip(rm_dies, rm_cells):
    df = df.mask((df["Die"] == die) & (df["Cell"] == cell))

stop = time.perf_counter()
loop_time = stop - start

# --

df = df_og.copy()
b100 = b100_og.copy()

start = time.perf_counter()

b100 = b100.drop_duplicates(["Die", "Cell"])
rm_dies = b100["Die"].to_list()
rm_cells = b100["Cell"].to_list()

for die, cell in zip(rm_dies, rm_cells):
    df = df.mask((df["Die"] == die) & (df["Cell"] == cell))

stop = time.perf_counter()
loop_no_dup_time = stop - start

# --

df = df_og.copy()
b100 = b100_og.copy()

start = time.perf_counter()

b100 = b100.drop_duplicates(["Die", "Cell"])

for die, cell in zip(b100["Die"], b100["Cell"]):
    df = df.mask((df["Die"] == die) & (df["Cell"] == cell))

stop = time.perf_counter()
loop_no_dup_zip_time = stop - start

# --

# INCORRECT
df = df_og.copy()
b100 = b100_og.copy()

start = time.perf_counter()

b100 = b100.drop_duplicates(["Die", "Cell"])

df = df.set_index(["Die", "Cell", "Current"])
df.loc[(b100["Die"], b100["Cell"], slice(None)), :] = np.nan

stop = time.perf_counter()
slice_time = stop - start
# INCORRECT

# --

df = df_og.copy()
b100 = b100_og.copy()

start = time.perf_counter()

df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))

df = df.mask(df["Die, Cell"].isin(b100["Die, Cell"]))

stop = time.perf_counter()
tuple_time = stop - start

# --

df = df_og.copy()
b100 = b100_og.copy()

start = time.perf_counter()

df["Die, Cell"] = list(zip(df["Die"], df["Cell"]))
b100["Die, Cell"] = list(zip(b100["Die"], b100["Cell"]))

df = df.mask(
    df[["Die", "Cell"]].apply(tuple, axis="columns").isin(
        b100[["Die", "Cell"]].apply(tuple, axis="columns")))

stop = time.perf_counter()
tuple_fly_time = stop - start

# --

print(f"Loop time: {loop_time:.2f} seconds")
print(f"Loop no dup time: {loop_no_dup_time:.2f} seconds")
print(f"Loop no dup zip time: {loop_no_dup_zip_time:.2f} seconds")
print(f"Slice time: {slice_time:.2f} seconds")
print(f"Tuple time: {tuple_time:.2f} seconds")
print(f"Tuple fly time: {tuple_fly_time:.2f} seconds")
```

```txt
Loop time: 14.78 seconds
Loop no dup time: 14.60 seconds
Loop no dup zip time: 15.54 seconds
Slice time: 0.16 seconds
Tuple time: 0.04 seconds
Tuple fly time: 0.31 seconds

Intel(R) Core(TM) i7-10510U CPU @ 1.80GHz
```

## Notes

Slicing is a close second on speed, but produces an incorrect result.
Same behavior as:
```py
df = df.mask((df["Die"].isin(b100["Die"])) & (df["Cell"].isin(b100["Cell"])))
```

The use of `drop_duplicates()` is technically not needed, since `b100`
is created at a single current, with no duplicate die, cell pairs.
Despite this, it is left in place for my own convenience.

---

Inspired by:  
https://stackoverflow.com/a/16068497  
https://stackoverflow.com/a/70355304