如何根据另一列中的值将函数应用于Pandas中的列？

Question

Thank you in advance for reading.

I have a dataframe:

df = pd.DataFrame({'Words':[{'Sec': ['level']},{'Sec': ['levels']},{'Sec': ['level']},{'Und': ['ba ']},{'Pro': ['conf'],'ProAbb': ['cth']}],'Conflict':[None,None,None,None,'Match Conflict']})


         Conflict                                     Words
0            None                      {u'Sec': [u'level']}
1            None                     {u'Sec': [u'levels']}
2            None                      {u'Sec': [u'level']}
3            None                        {u'Und': [u'ba ']}
4  Match Conflict  {u'ProAbb': [u'cth'], u'Pro': [u'conf']}

I want to apply a routine that, for each element in 'Words' , checks if Conflict = 'Match Conflict' and if so, applies some function to the value in 'Words' .

For instance, using the following placeholder function:

def func(x):
    x = x.clear()
    return x

I write:

df['Words'] = df[df['Conflict'] == 'Match Conflict']['Words'].apply(lambda x: func(x))

My expected output is:

         Conflict                                     Words
0            None                      {u'Sec': [u'level']}
1            None                     {u'Sec': [u'levels']}
2            None                      {u'Sec': [u'level']}
3            None                        {u'Und': [u'ba ']}
4  Match Conflict                                        None

Instead I get:

         Conflict Words
0            None   NaN
1            None   NaN
2            None   NaN
3            None   NaN
4  Match Conflict  None

The function is applied only to the row which has Conflict = 'Match Conflict' but at the expense of the other rows (which all become None . I assumed the other rows would be left untouched; obviously this is not the case.

Can you explain how I might achieve my desired output without dropping all of the information in the Words column? I believe the answer may lie with np.where but I have not been able to make this work, this was the best I could come up with.

Any help much appreciated. Thanks.

Answer 1

You should rewrite the function to work with all of your rows:

def func(x, match):
    if x['Conflict'] == match:
        return None
    return x['Words']

df['Words'] = df.apply(lambda row: func(row, 'Match Conflict'), axis=1)

Answer 2

You can try to update only those rows that match the condition using .loc :

df.loc[df['Conflict'] == 'Match Conflict', 'Words'] = df.loc[df['Conflict'] == 'Match Conflict', 'Words'].apply(lambda x: func(x))

Answer 3

You can also use where as you described,

condition = df.Conflict != 'Match Conflict'
df['Words'] = df.Words.where(condition, None)

         Conflict                  Words
0            None   {u'Sec': [u'level']}
1            None  {u'Sec': [u'levels']}
2            None   {u'Sec': [u'level']}
3            None     {u'Und': [u'ba ']}
4  Match Conflict                   None

Answer 4

suppose a placeholder

def func(x):
    x = x.clear()
    return x

Then we can use boolean indexing and apply to obtain the desired output.

df.ix[df['Conflict']=='Match Conflict', 'Words'].apply(func)

I wanted to provide a concise one-liner but I was too late :,(