My dataframe reads like :
df1
user_id username firstname lastname
123 abc abc abc
456 def def def
789 ghi ghi ghi
df2
user_id username firstname lastname
111 xyz xyz xyz
456 def def def
234 mnp mnp mnp
Now I want a output dataframe like
user_id username firstname lastname
123 abc abc abc
456 def def def
789 ghi ghi ghi
111 xyz xyz xyz
234 mnp mnp mnp
As user_id 456
is common across both the dataframes. I have tried groupby on user_id groupby(['user_id'])
. But looks like groupby need to be followed by some aggregation
which I don't want here.
Use concat
+ drop_duplicates
:
df = pd.concat([df1, df2]).drop_duplicates('user_id').reset_index(drop=True)
print (df)
user_id username firstname lastname
0 123 abc abc abc
1 456 def def def
2 789 ghi ghi ghi
3 111 xyz xyz xyz
4 234 mnp mnp mnp
Solution with groupby
and aggregate first
is slowier:
df = pd.concat([df1, df2]).groupby('user_id', as_index=False, sort=False).first()
print (df)
user_id username firstname lastname
0 123 abc abc abc
1 456 def def def
2 789 ghi ghi ghi
3 111 xyz xyz xyz
4 234 mnp mnp mnp
EDIT:
Another solution with boolean indexing
and numpy.in1d
:
df = pd.concat([df1, df2[~np.in1d(df2['user_id'], df1['user_id'])]], ignore_index=True)
print (df)
user_id username firstname lastname
0 123 abc abc abc
1 456 def def def
2 789 ghi ghi ghi
3 111 xyz xyz xyz
4 234 mnp mnp mnp
One approach with masking -
def app1(df1,df2):
df20 = df2[~df2.user_id.isin(df1.user_id)]
return pd.concat([df1, df20],axis=0)
Two more approaches using the underlying array data, np.in1d
, np.searchsorted
to get the mask of matches and then stacking those two and constructing an output dataframe from the stacked array data -
def app2(df1,df2):
df20_arr = df2.values[~np.in1d(df1.user_id.values, df2.user_id.values)]
arr = np.vstack(( df1.values, df20_arr ))
df_out = pd.DataFrame(arr, columns= df1.columns)
return df_out
def app3(df1,df2):
a = df1.values
b = df2.values
df20_arr = b[~np.in1d(a[:,0], b[:,0])]
arr = np.vstack(( a, df20_arr ))
df_out = pd.DataFrame(arr, columns= df1.columns)
return df_out
def app4(df1,df2):
a = df1.values
b = df2.values
b0 = b[:,0].astype(int)
as0 = np.sort(a[:,0].astype(int))
df20_arr = b[as0[np.searchsorted(as0,b0)] != b0]
arr = np.vstack(( a, df20_arr ))
df_out = pd.DataFrame(arr, columns= df1.columns)
return df_out
Timings for given sample -
In [49]: %timeit app1(df1,df2)
...: %timeit app2(df1,df2)
...: %timeit app3(df1,df2)
...: %timeit app4(df1,df2)
...:
1000 loops, best of 3: 753 µs per loop
10000 loops, best of 3: 192 µs per loop
10000 loops, best of 3: 181 µs per loop
10000 loops, best of 3: 171 µs per loop
# @jezrael's edited solution
In [85]: %timeit pd.concat([df1, df2[~np.in1d(df2['user_id'], df1['user_id'])]], ignore_index=True)
1000 loops, best of 3: 614 µs per loop
Would be interesting to see how these fare on larger datasets.
Another approach is to use np.in1d to check for duplicate user_id.
pd.concat([df1,df2[df2.user_id.isin(np.setdiff1d(df2.user_id,df1.user_id))]])
Or to use a set to get unique rows from the merged records from df1 and df2. This one seems to be a few times faster.
pd.DataFrame(data=np.vstack({tuple(row) for row in np.r_[df1.values,df2.values]}),columns=df1.columns)
Timings:
%timeit pd.concat([df1,df2[df2.user_id.isin(np.setdiff1d(df2.user_id,df1.user_id))]])
1000 loops, best of 3: 2.48 ms per loop
%timeit pd.DataFrame(data=np.vstack({tuple(row) for row in np.r_[df1.values,df2.values]}),columns=df1.columns)
1000 loops, best of 3: 632 µs per loop