I have two dataframes; the first one(df1) is:
df1 = pd.DataFrame({'col1': [0,1], 'col2': [0,1]})
df1 = df1.rename(index = {k:v for k,v in zip([0,1],['zero','one'])})
print(df1)
col1 col2
zero 0 0
one 1 1
and the second one(df2) is:
df2 = pd.DataFrame({k:v for k,v in zip(['col'+str(i) for i in range(3,10)],
[[2,3]]*7)
})
print(df2)
col3 col4 col5 col6 col7 col8 col9
0 2 2 2 2 2 2 2
1 3 3 3 3 3 3 3
the final product(df3) should look exactly like:
col1 col2 col3 col4 col5 col6 col7 col8 col9
zero 0 0 2 2 2 2 2 2 2
one 1 1 3 3 3 3 3 3 3
This is the way I do it, which is a bit unpythonic to my taste:
df3 = df1.reset_index(drop = True)
df3 = df3.join(df2.reset_index(drop = True))
df3 = df3.rename(index = {k:v for k,v in zip(df3.index,df1.index)})
print(df3)
Is there any one-line code that can do the job? Thank you guys
You can create df2.index
by df1.index
by set_index
, only necessary same length of both DataFrames
:
df = df1.join(df2.set_index(df1.index))
Or:
df = pd.concat([df1, df2.set_index(df1.index)], axis=1)
print (df)
col1 col2 col3 col4 col5 col6 col7 col8 col9
zero 0 0 2 2 2 2 2 2 2
one 1 1 3 3 3 3 3 3 3
If have list same length like both DataFrames pass nested list for distinguish you want pass list, not list of column names (df2.set_index(L)
or df2.set_index(['a','b'])
):
L = ['a','b']
df = pd.concat([df1.set_index([L]), df2.set_index([L])], axis=1)
print (df)
col1 col2 col3 col4 col5 col6 col7 col8 col9
a 0 0 2 2 2 2 2 2 2
b 1 1 3 3 3 3 3 3 3
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