Q&A

# Python looping 300 000 rows

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Based on my last question comes new one.
How to loop over 300 000 rows and edit each row string one by one? I have a list of 11-digit numbers stored in one single column in Excel, and I need to separate the digits according to this pattern: `2-2-1-3-3`.

I use the code below to loop to test the solution for only 20 rows and it's working.

Example: `00002451018` becomes `00 00 2 451 018`.

`priceListTest` contains the column `Column1` which has these 11 digit numbers. Somehow I need to loop all over these 300 000 rows and use the `get_slices` to change the pattern for each row like from the example above and store it into the new column `New Value`.

The `for index, row` it's working very slowly when I have to use it for 300 000 rows. Maybe there is a better method, but I'm new to python.

``````for index, row in priceListTest.iterrows():
#print(index,row)
def get_slices(n, sizes, n_digits=11):
for size in sizes:
n_digits -= size

val, n = divmod(n, 10 ** n_digits)
yield f'{val:0{size}}'

n = row['Column1']
newVar = (' '.join(get_slices(n, [2, 2, 1, 3, 3])))
priceListTest.at[index,['New Value']] = newVar``````
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Parallel execution (1 comment)
The actual performance issue (5 comments)
Types (1 comment)
Create the function just once (3 comments)
A small note regarding MCVE (1 comment)

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I’m struggling to get `timeit` working correctly but this is faster in my limited tests:

``````l = [123456789, 23456789012, 34567890123]

result = [0, 0, 0]

for idx, row in enumerate(l):
i = f"{row:011}"
result[idx] = f"{i[:2]}-{i[2:4]}-{i[4:5]}-{i[5:8]}-{i[8:]}"

print(result)
``````
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