python - How to avoid mutiple columns on Pandas.Merge -


imagine have following dataframes on pandas:

in [7]: a= pd.dataframe([['foo'],['bar'],['quz'],['baz']],columns=['key'])  in [8]: a['value'] = 'none'  in [9]: out[9]:    key value 0  foo  none 1  bar  none 2  quz  none 3  baz  none  in [10]: b = pd.dataframe([['foo',5],['bar',6],['quz',7]],columns= ['key','value'])  in [11]: b out[11]:    key  value 0  foo      5 1  bar      6 2  quz      7  in [12]: pd.merge(a,b, on='key', how='outer') out[12]:    key value_x  value_y 0  foo    none        5 1  bar    none        6 2  quz    none        7 3  baz    none      nan

but want (avoiding repeat column basically):

   key  value 0  foo  5 1  bar  6 2  quz  7 3  baz  nan

i suppose can take output , drop _x value , rename _y seems overkill. on sql trivial.

edit:

john recomended use:

in [1]: a.set_index('key', inplace=true)         a.update(b.set_index('key'), join='left', overwrite=true)         a.reset_index(inplace=true)

this works , asked for.

in example merging 2 dataframes same column, 1 contains strings ('none') other integers, pandas doesn't know column value want keep , should replaced, creates column both.

you can use update instead

in [10]: a.update(b, join='left', overwrite=true) in [11]: out[11]:      key value 0   foo 5 1   bar 6 2   quz 7 3   baz nan

another solution state values want given column:

in [15]: a.loc[b.index, 'value'] = b.value in [16]: out[16]:      key value 0   foo 5 1   bar 6 2   quz 7 3   baz nan

personally prefer second solution because know happening, first closer looking in question.

edit:

if indices don't match, i'm not quite sure how make happen. hence suggest making them match:

in [1]: a.set_index('key', inplace=true)         a.update(b.set_index('key'), join='left', overwrite=true)         a.reset_index(inplace=true)

it may there better way this, don't believe pandas has way perform operation outright.

the second solution can used updated index:

in [24]: a.set_index('key', inplace=true)          a.loc[b.key, 'value'] = b.value.tolist()

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