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Select rows of dataframe whose column values amount to a given sum

Time:01-18

I need to find out how many of the first N rows of a dataframe make up (just over) 50% of the sum of values for that column.

Here's an example:

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.rand(10, 1), columns=list("A"))

0  0.681991
1  0.304026
2  0.552589
3  0.716845
4  0.559483
5  0.761653
6  0.551218
7  0.267064
8  0.290547
9  0.182846

therefore

sum_of_A = df["A"].sum()

4.868260213425804

and with this example I need to find, starting from row 0, how many rows I need to get a sum of at least 2.43413 (approximating 50% of sum_of_A).

Of course I could iterate through the rows and sum and break when I get over 50%, but is there a more concise/Pythonic/efficient way of doing this?

CodePudding user response:

I would use .cumsum(), which we can use to get all the rows where the cumulative sum is at least half of the total sum:

df[df["A"].cumsum() < df["A"].sum() / 2]
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