2

I'm trying to update the pandas data frame by logical condition but, it fails with below error,

df[df.b <= 0]['b'] = 0

A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead

How do I get this working?

Data:

df = pd.DataFrame({'a': np.random.randn(4), 'b': np.random.randn(4)})
    a           b
0   1.462028    -1.337630
1   0.206193    -1.060710
2   -0.464847   -1.881426
3   0.290627    0.650805

I am learning pandas. In R, syntax is like below,

df[df$b <= 0]$b <- 0
Improve this question

3 Answers 3

Reset to default
7

Use 

df.loc[df.b <= 0, 'b']= 0

For efficiency pandas just creates a references from the previous DataFrame instead of creating new DataFrame every time a filter is applied.
Thus when you assign a value to DataFrame it needs tobe updated in the source DataFrame (not just the current slice of it). This is what is refered in the warning

A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead

To avoid this .loc syntax is used.

For more information on DataFrame indexing

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

Comments

0

Try this:

>>> df.ix[df['b']<=0] = 0
>>> df
      a         b
0  0.000000  0.000000
1  0.000000  0.000000
2  0.212535  0.491969
3  0.000000  0.000000

Note: Since v0.20 ix has been deprecated. Use loc or iloc instead.

Improve this answer

Comments

0

Follow below pattern for updating the value - 

food_reviews_df.loc[food_reviews_df.Score <= 3, 'Score'] = 0
food_reviews_df.loc[food_reviews_df.Score >= 4, 'Score'] = 1
Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.