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 
                To get the nth row in a Pandas DataFrame, we can use the iloc() method. For example, df. iloc[4] will return the 5th row because row numbers start from 0.
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) 
                        If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With