This is solved in 2 steps:

• Find rows matching remove conditions
• Do anti-left-join on the the composite key (Die, Cell)

To filter out the rows:

# Read in the data
df = pd.read_csv("data.csv")

# Identify cells with low current
b100 = df[(df["Resistance"] < 100000) & (df["Current"] == 100)]

# Pull out only columns of interest
bad_cells = b100[["Die", "Cell"]]

We can take unique values of bad_cells here, but I didn't, because it doesn't affect the outcome in the end. Also, taking the column subset is not necessary, but reduces extra junk columns later.

Pandas has no explicit anti-join ("join on doesn't equal") so I stole it from https://stackoverflow.com/a/55543744/21703684:

# Remove them with anti join
outer_join = df.merge(bad_cells, how="left", indicator=True)
filtered = outer_join[outer_join._merge != "both"].drop("_merge", axis=1)

print(filtered)

posted over 1 year ago

CC BY-SA 4.0

matthewsnyder‭

2209 reputation 52 59 259 92

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.

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"]))

    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:

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")

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):

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

posted over 1 year ago

CC BY-SA 4.0

1y ago

mcp‭

152 reputation 8 2 21 23

You were on the right path breaking it down into two steps. First identify the cells you want to remove:

# Identify the 'Cell' values that meet the criteria
cells_to_remove = df[(df['Resistance'] < 100000) & (df['Current'] == 100)]['Cell'].unique()

And then remove them:

# Remove all rows with 'Cell' values that were identified
filtered_df = df[~df['Cell'].isin(cells_to_remove)]

posted over 1 year ago

CC BY-SA 4.0

dagelf‭

29 reputation 0 2 2 0