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Скачать или смотреть Flag the max value in each column of a DataFrame as True and the rest as False

  • Emrah KAYA
  • 2025-08-15
  • 0
Flag the max value in each column of a DataFrame as True and the rest as False
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Hello everyone! I hope this video has helped solve your questions and issues. This video is shared because a solution has been found for the question/problem. I create videos for questions that have solutions. If you have any other issues, feel free to reach out to me on Instagram:   / ky.emrah  

Below, you can find the text related to the question/problem. In the video, the question will be presented first, followed by the answers. If the video moves too fast, feel free to pause and review the answers. If you need more detailed information, you can find the necessary sources and links at the bottom of this description. I hope this video has been helpful, and even if it doesn't directly solve your problem, it will guide you to the source of the solution. I'd appreciate it if you like the video and subscribe to my channel!Flag the max value in each column of a DataFrame as True and the rest as False

I have a DataFrame that I am rounding. After the round, I subtract the original from the resultant. This gives me a data frame with a shape identical to the original, but which contains the amount of change the rounding operation caused.
I need to transform this into a Boolean where there is a true flag for the max of the row, and everything else in the row is false. All steps but the final one are handled with a vectorized function. But I can't seem to figure out how to vectorize the last step. This is what I am currently doing:
a = pd.DataFrame([[2.290119, 5.300725, 17.266693, 75.134857, 0.000000, 0.000000, 0.007606],
[0.000000, 7.560276, 55.579175, 36.858266, 0.000000, 0.000000, 0.002284],
[0.001574, 15.225538, 39.309742, 45.373800, 0.000951, 0.001198, 0.087197],
[0.000000, 55.085390, 15.547927, 29.327661, 0.000000, 0.017691, 0.021331],
[0.000000, 66.283488, 15.636673, 17.912315, 0.000000, 0.003185, 0.164339]])

b = a.round(-1) # round to 10's place (not 10ths)
c = b-a
round_modifier = c.apply(lambda x: x.eq(x.max()), axis="columns")
print(round_modifier)

a = pd.DataFrame([[2.290119, 5.300725, 17.266693, 75.134857, 0.000000, 0.000000, 0.007606],
[0.000000, 7.560276, 55.579175, 36.858266, 0.000000, 0.000000, 0.002284],
[0.001574, 15.225538, 39.309742, 45.373800, 0.000951, 0.001198, 0.087197],
[0.000000, 55.085390, 15.547927, 29.327661, 0.000000, 0.017691, 0.021331],
[0.000000, 66.283488, 15.636673, 17.912315, 0.000000, 0.003185, 0.164339]])

b = a.round(-1) # round to 10's place (not 10ths)
c = b-a
round_modifier = c.apply(lambda x: x.eq(x.max()), axis="columns")
print(round_modifier)

0 1 2 3 4 5 6
0 False False False True False False False
1 False False True False False False False
2 False True False False False False False
3 False True False False False False False
4 False False True False False False False

0 1 2 3 4 5 6
0 False False False True False False False
1 False False True False False False False
2 False True False False False False False
3 False True False False False False False
4 False False True False False False False

I am aware of DataFrame.idxmax(axis="columns"), which gives me the column name (of each row) where the max is found, but I can't seem to find a (pythonic) way to take that and populate the corresponding flag with a True. The lambda expression I'm using gives the correct result, but I'm hoping for a faster method.
DataFrame.idxmax(axis="columns")
For anyone wondering, the use case is that I want to round the values in the original data frame to the tens place, such that they sum to 100. I have pre-scaled this data so it should be close, but the rounding can cause the sum to come to 90 or 110. I intend to use this T/F matrix to decide which rounded value caused the most delta, then round it in the opposite direction since this is the minimum impact method with which to coerce the series to properly sum to 100 in chunks of 10.


Tags: python,pandas,dataframe,max,roundingSource of the question:
https://stackoverflow.com/questions/7...

Question and source license information:
https://meta.stackexchange.com/help/l...
https://stackoverflow.com/

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