44

I have a DataFrame and I would select only rows that contain index value into df1.index.

for Example:

In [96]: df
Out[96]:
   A  B  C  D
1  1  4  9  1
2  4  5  0  2
3  5  5  1  0
22 1  3  9  6

and these indexes

In[96]:df1.index
Out[96]:
Int64Index([  1,   3,   4,   5,   6,   7,  22,  28,  29,  32,], dtype='int64', length=253)

I would like this output:

In [96]: df
Out[96]:
   A  B  C  D
1  1  4  9  1
3  5  5  1  0
22 1  3  9  6
Henry Ecker
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giupardeb
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2 Answers2

75

Use isin:

df = df[df.index.isin(df1.index)]

Or get all intersectioned indices and select by loc:

df = df.loc[df.index & df1.index]
df = df.loc[np.intersect1d(df.index, df1.index)]
df = df.loc[df.index.intersection(df1.index)]

print (df)
    A  B  C  D
1   1  4  9  1
3   5  5  1  0
22  1  3  9  6

EDIT:

I tried solution: df = df.loc[df1.index]. Do you think that this solution is correct?

Solution is incorrect:

df = df.loc[df1.index]
print (df)

      A    B    C    D
1   1.0  4.0  9.0  1.0
3   5.0  5.0  1.0  0.0
4   NaN  NaN  NaN  NaN
5   NaN  NaN  NaN  NaN
6   NaN  NaN  NaN  NaN
7   NaN  NaN  NaN  NaN
22  1.0  3.0  9.0  6.0
28  NaN  NaN  NaN  NaN
29  NaN  NaN  NaN  NaN
32  NaN  NaN  NaN  NaN
C:/Dropbox/work-joy/so/_t/t.py:23: FutureWarning: 
Passing list-likes to .loc or [] with any missing label will raise
KeyError in the future, you can use .reindex() as an alternative.

See the documentation here:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#deprecate-loc-reindex-listlike
  print (df)
jezrael
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10

Passing the index to the row indexer/slicer of .loc now works, you just need to make sure to specify the columns as well, i.e.:

df = df.loc[df1.index, :]  # works

and NOT

df = df.loc[df1.index] # won't work

IMO This is more neater/consistent with the expected usage of .loc

Hansang
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  • You're right, it seems that the devs have changed the implementation. However now both seem to raise the KeyError warning if you pass in a list of values that doesn't exist – Hansang Jan 20 '20 at 10:45
  • 2
    This solution has the advantage of ensuring order of df and df1 are the same. – DocOc Sep 18 '21 at 16:24
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    this only works if `df1`'s entire index is contained in `df`'s index; the accepted answer does not have that limitation – fantabolous Jun 09 '23 at 04:31