How to use lists containing row/column indices to replace several single values of dataframe?

Question

Look at the following code:

import numpy as np
import pandas as pd

np.random.seed(42)
arr = np.random.randint(1, 11, (5, 5)).astype(np.float)
df = pd.DataFrame(arr)

>>> df
Out[49]: 
      0    1     2    3    4
0   7.0  4.0   8.0  5.0  7.0
1  10.0  3.0   7.0  8.0  5.0
2   4.0  8.0   8.0  3.0  6.0
3   5.0  2.0   8.0  6.0  2.0
4   5.0  1.0  10.0  6.0  9.0

rows, cols = [0, 2], [1, 3]
arr[rows, cols] = np.nan
df = pd.DataFrame(arr)

>>> df
Out[50]: 
      0    1     2    3    4
0   7.0  NaN   8.0  5.0  7.0
1  10.0  3.0   7.0  8.0  5.0
2   4.0  8.0   8.0  NaN  6.0
3   5.0  2.0   8.0  6.0  2.0
4   5.0  1.0  10.0  6.0  9.0

Is there a way to replace the values at the defined indices with e.g. NaN that does not require the conversion of the dataframe to a numpy array?

Are you only interested in replacing nan values? This post will should be useful if so. — busybear
– busybear, Commented Jan 30, 2019 at 18:45
Thanks for the response! No Im interested in a numpy-like replacement using the indices of rows & columns (not index and column names, but the numerical indices) — zwithouta
– zwithouta, Commented Jan 30, 2019 at 19:25

Colonder · Accepted Answer · 2019-01-30 14:10:04Z

1

Try this:

df.values[rows, cols]

answered Jan 30, 2019 at 14:10

Colonder

1,5763 gold badges21 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

busybear Over a year ago

values will actually give a numpy array.

zwithouta Over a year ago

This is the shortcut I have searched for. df.values outputs an numpy array, but df.values[rows, cols] = np.nan updates the dataframe! This allows me to replace values in the dataframe by using numerical indices (that is I don't have to care about the labeling of the rows/columns of the dataframe)

BENY · Accepted Answer · 2019-01-30 15:20:49Z

0

Using lookup

df.lookup(rows,cols)
Out[810]: array([4, 3])

answered Jan 30, 2019 at 15:20

BENY

324k22 gold badges176 silver badges250 bronze badges

3 Comments

Colonder Over a year ago

This is essentially doing the same things as in my answer, or iterating over the indices using zip() and getting each value at a time

BENY Over a year ago

@Colonder not always same , this one is index and columns base, .values is position base , so close but not same .

zwithouta Over a year ago

Thanks for the reply. I edited the question a little bit. @Colonder indeed came up with what I was looking for

Collectives™ on Stack Overflow

How to use lists containing row/column indices to replace several single values of dataframe?

2 Answers 2

2 Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related