如何检查一个数据帧中的列值是否可用或不检查另一数据帧的列中的值？

Question

I have two dataframes-

df1_data = {'sym1' :{0:'abc a01',1:'pqr q02',2:'xyz y03',3:'mno o12',4:'lmn l45'}}
df1 = pd.DataFrame(df1_data)
print df1

df2_data = {'sym2' :{0:'abc a01',1:'xxx p0',2:'xyz y03',3:'mno o12',4:'lmn l45',5:'rrr r1',6:'kkk k3'}}
df2 = pd.DataFrame(df2_data)
print df2

output-

      sym1
0  abc a01
1  pqr q02
2  xyz y03
3  mno o12
4  lmn l45
      sym2
0  abc a01
1   xxx p0
2  xyz y03
3  mno o12
4  lmn l45
5   rrr r1
6   kkk k3

I want to check sym2 column values available or not in df2 dataframes sym1 column. If symbols in sym2 column are not available then I want list of that symbols which are not available in sym1 column. If all symbols are available then list must be empty.

Expected Result-

list -> ['xxx p0','rrr r1','kkk k3']

Answer 1

You can use boolean indexing with isin , then select by ix and convert to list by tolist :

print (~df2.sym2.isin(df1.sym1))
0    False
1     True
2    False
3    False
4    False
5     True
6     True
Name: sym2, dtype: bool

print (df2.ix[~df2.sym2.isin(df1.sym1), 'sym2'])
1    xxx p0
5    rrr r1
6    kkk k3
Name: sym2, dtype: object

print (df2.ix[~df2.sym2.isin(df1.sym1), 'sym2'].tolist())
['xxx p0', 'rrr r1', 'kkk k3']

Answer 2

Here is another, bit faster, solution:

In [54]: df2.set_index('sym2').index.difference(df1.set_index('sym1').index).values
Out[54]: array(['kkk k3', 'rrr r1', 'xxx p0'], dtype=object)

or as vanilla Python list:

In [74]: df2.set_index('sym2').index.difference(df1.set_index('sym1').index).values.tolist()
Out[74]: ['kkk k3', 'rrr r1', 'xxx p0']

Timings for 700K and 500K DFs:

In [55]: df1 = pd.concat([df1] * 10**5, ignore_index=True)

In [57]: df2 = pd.concat([df2] * 10**5, ignore_index=True)

In [58]: df1.shape
Out[58]: (500000, 1)

In [59]: df2.shape
Out[59]: (700000, 1)

In [67]: %timeit df2.set_index('sym2').index.difference(df1.set_index('sym1').index).values
10 loops, best of 3: 123 ms per loop

In [68]: %timeit df2.ix[~df2.sym2.isin(df1.sym1), 'sym2']
1 loop, best of 3: 216 ms per loop

In [72]: %timeit df2.set_index('sym2').index.difference(df1.set_index('sym1').index).values.tolist()
10 loops, best of 3: 123 ms per loop