Correct way to set new column in pandas DataFrame to avoid SettingWithCopyWarning

Question:

Trying to create a new column in the netc df but i get the warning

netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

C:Anacondalibsite-packagesipykernel__main__.py:1: SettingWithCopyWarning: 
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

whats the proper way to create a field in the newer version of Pandas to avoid getting the warning?

pd.__version__
Out[45]:
u'0.19.2+0.g825876c.dirty'
Asked By: Daniel

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Answers:

As it says in the error, try using .loc[row_indexer,col_indexer] to create the new column.

netc.loc[:,"DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM.

Notes

By the Pandas Indexing Docs your code should work.

netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

gets translated to

netc.__setitem__('DeltaAMPP', netc.LOAD_AM - netc.VPP12_AM)

Which should have predictable behaviour. The SettingWithCopyWarning is only there to warn users of unexpected behaviour during chained assignment (which is not what you’re doing). However, as mentioned in the docs,

Sometimes a SettingWithCopy warning will arise at times when there’s no obvious chained indexing going on. These are the bugs that SettingWithCopy is designed to catch! Pandas is probably trying to warn you that you’ve done this:

The docs then go on to give an example of when one might get that error even when it’s not expected. So I can’t tell why that’s happening without more context.

Answered By: Filip Kilibarda

Your example is incomplete, as it doesn’t show where netc comes from. It is likely that netc itself is the product of slicing, and as such Pandas cannot make guarantees that it isn’t a view or a copy.

For example, if you’re doing this:

netc = netb[netb["DeltaAMPP"] == 0]
netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

then Pandas wouldn’t know if netc is a view or a copy. If it were a one-liner, it would effectively be like this:

netb[netb["DeltaAMPP"] == 0]["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

where you can see the double indexing more clearly.

If you want to make netc separate from netb, one possible remedy might be to force a copy in the first line (the loc is to make sure we’re not copying two times), like:

netc = netb.loc[netb["DeltaAMPP"] == 0].copy()

If, on the other hand, you want to have netb modified with the new column, you may do:

netb.loc[netb["DeltaAMPP"] == 0, "DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM
Answered By: Ronan Paixão

You need to reset_index when you will create column especially if you have filtered on specific values… then you don’t need to use .loc[row_indexer,col_indexer]

netc.reset_index(drop=True, inplace=True)
netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

Then it should work 🙂

Answered By: Mohamed T. Bouzid

I had the SettingWithCopyWarning-issue, when assigning data to a DataFrame df, which was constructed by indexing.
Both commands

  • df['new_column'] = something
  • df.loc[:, 'new_column'] = something

did not work without the warning.
As soon as copying df (DataFrame.copy()) everything was fine.

In the code below, compare df0 = df_test[df_test['a']>3] and df1 = df_test[df_test['a']>3].copy().
For df0 both assignments throw the Warning. For df1 both work fine.

>>> df_test
      a     b     c     d  e
0   0.0   1.0   2.0   3.0  0
1   4.0   5.0   6.0   7.0  1
2   8.0   9.0  10.0  11.0  2
3  12.0  13.0  14.0  15.0  3
4  16.0  17.0  18.0  19.0  4
>>> df0 = df_test[df_test['a']>3]
>>> df1 = df_test[df_test['a']>3].copy()
>>> df0['e'] = np.arange(4)
__main__:1: SettingWithCopyWarning: 
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

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
>>> df1['e'] = np.arange(4)
>>> df0.loc[2, 'a'] = 77
/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py:1719: SettingWithCopyWarning: 
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

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  self._setitem_single_column(loc, value, pi)
>>> df1.loc[2, 'a'] = 77
>>> df0
      a     b     c     d  e
1   4.0   5.0   6.0   7.0  0
2  77.0   9.0  10.0  11.0  1
3  12.0  13.0  14.0  15.0  2
4  16.0  17.0  18.0  19.0  3
>>> df1
      a     b     c     d  e
1   4.0   5.0   6.0   7.0  0
2  77.0   9.0  10.0  11.0  1
3  12.0  13.0  14.0  15.0  2
4  16.0  17.0  18.0  19.0  3

By the way: It is recommended to read the docs about this issue (Link in Warning)

Answered By: Markus Dutschke

As pointed out in other answers, there is a good chance that you have done some filtering on the data, else this warning should not have popped up (since your steps are correct).

Assuming you have done some filtering, you could try doing the following steps:

netc_copied = netc.copy()
netc.loc[:, "DeltaAMPP"] = netc_copied["LOAD_AM"] - netc_copied["VPP12_AM"]

Note that I have added a new column in the original DataFrame. You could do this in the copied DataFrame too.

Answered By: ankitgupta5294

A simpler solution is to just use ‘assign’:

netc = netc.assign(DeltaAMPP=netc_copied['LOAD_AM']-netc_copied['VPP12_AM'])
Answered By: AbhiGupta

Alternatively you can also use eval:

netc.eval('DeltaAMPP = LOAD_AM - VPP12_AM', inplace = True)

Since inplace=True you don’t need to assign it back to netc.

Answered By: rachwa

you’ll will still get error even after using .loc or .iloc for slicing, all you have to do is reset index after slicing

df.reset_index()
Answered By: Naresh Kumar Sekar
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