Concatenate cells into a string with separator pandas python

Question

Given the following:

df = pd.DataFrame({\'col1\' : [\"a\",\"b\"],
            \'col2\'  : [\"ab\",np.nan], \'col3\' : [\"w\",\"e\"]})

I would l

Answer 1

In [1556]: df.apply(lambda x: '*'.join(x.dropna().astype(str).values), axis=1)
Out[1556]: 
0    a*ab*w
1       b*e
2     3*4*�
3     ñ*ü*á
dtype: object

Answer 2

df.apply(lambda row: '*'.join(row.dropna()), axis=1)

Answer 3

In [68]:

df['new_col'] = df.apply(lambda x: '*'.join(x.dropna().values.tolist()), axis=1)
df
Out[68]:
  col1 col2 col3 new_col
0    a   ab    w  a*ab*w
1    b  NaN    e     b*e

UPDATE

If you have ints or float you can convert these to str first:

In [74]:

df = pd.DataFrame({'col1' : ["a","b",3],
            'col2'  : ["ab",np.nan, 4], 'col3' : ["w","e", 6]})
df
Out[74]:
  col1 col2 col3
0    a   ab    w
1    b  NaN    e
2    3    4    6
In [76]:

df['new_col'] = df.apply(lambda x: '*'.join(x.dropna().astype(str).values), axis=1)
df
Out[76]:
  col1 col2 col3 new_col
0    a   ab    w  a*ab*w
1    b  NaN    e     b*e
2    3    4    6   3*4*6

Another update

In [81]:

df = pd.DataFrame({'col1' : ["a","b",3,'ñ'],
            'col2'  : ["ab",np.nan, 4,'ü'], 'col3' : ["w","e", 6,'á']})
df
Out[81]:
  col1 col2 col3
0    a   ab    w
1    b  NaN    e
2    3    4    6
3    ñ    ü    á

In [82]:

df['new_col'] = df.apply(lambda x: '*'.join(x.dropna().astype(str).values), axis=1)
​
df
Out[82]:
  col1 col2 col3 new_col
0    a   ab    w  a*ab*w
1    b  NaN    e     b*e
2    3    4    6   3*4*6
3    ñ    ü    á   ñ*ü*á

My code still works with Spanish characters

Answer 4

You can use dropna()

df['col4'] = df.apply(lambda row: '*'.join(row.dropna()), axis=1)

UPDATE:

Since, you need to convert numbers and special chars too, you can use astype(unicode)

In [37]: df = pd.DataFrame({'col1': ["a", "b"], 'col2': ["ab", np.nan], "col3": [3, u'\xf3']})

In [38]: df.apply(lambda row: '*'.join(row.dropna().astype(unicode)), axis=1)
Out[38]: 
0    a*ab*3
1       b*ó
dtype: object

In [39]: df['col4'] = df.apply(lambda row: '*'.join(row.dropna().astype(unicode)), axis=1)

In [40]: df
Out[40]: 
  col1 col2 col3    col4
0    a   ab    3  a*ab*3
1    b  NaN    ó     b*ó

Answer 5

for row in xrange(len(df)):
    s = '*'.join(df.ix[row].dropna().tolist())
    print s