How to drop Ith row of a data frame

Question

How do I drop row number i of a DF ?

I did the thing below but it is not working.

 DF = DF.drop(i)

So I wonder what I miss there.

Asked By: TourEiffel

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Source

Answer 1

You need to add square brackets:

df = df.drop([i])

Answered By: Lorenzo Bonetti

Answer 2

You must pass a label to drop. Here drop tries to use i as a label and fails (ith KeyError) as your index probably has other values. Worse, if the index was composed of integers in random order you might drop an incorrect row without noticing it.

Use:

df.drop(df.index[i])

Example:

df = pd.DataFrame({'col': range(4)}, index=list('ABCD'))
out = df.drop(df.index[2])

output:

   col
A    0
B    1
D    3

pitfall

In case of duplicated indices, you might remove unwanted rows!

df = pd.DataFrame({'col': range(4)}, index=list('ABAD'))
out = df.drop(df.index[2])

output (A is incorrectly dropped!):

   col
B    1
D    3

workaround:

import numpy as np

out = df[np.arange(len(df)) != i]

drop several indices by position:

import numpy as np

out = df[~np.isin(np.arange(len(df)), [i, j])]

Answered By: mozway

Answer 3

Try This:

df.drop(df.index[i])

Answered By: Galaxy

How to drop Ith row of a data frame

Question:

Answers:

pitfall

workaround: