pandas 使用分隔符pandas python将单元格连接成一个字符串
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Concatenate cells into a string with separator pandas python
提问by Bastien
Given the following:
鉴于以下情况：
df = pd.DataFrame({'col1' : ["a","b"],
'col2' : ["ab",np.nan], 'col3' : ["w","e"]})
I would like to be able to create a column that joins the content of all three columns into one string, separated by the character "*" while ignoring NaN.
我希望能够创建一个列，将所有三列的内容连接成一个字符串，由字符“*”分隔，同时忽略NaN.
so that I would get something like that for example:
这样我就会得到类似的东西，例如：
a*ab*w
b*e
Any ideas?
有任何想法吗？
Just realised there were a few additional requirements, I needed the method to work with ints and floats and also to be able to deal with special characters (e.g., letters of Spanish alphabet).
刚刚意识到有一些额外的要求，我需要这种方法来处理整数和浮点数，还需要能够处理特殊字符（例如，西班牙字母表的字母）。
回答by EdChum
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
更新
If you have ints or float you can convert these to strfirst:
如果您有整数或浮点数，您可以str先将它们转换为：
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
另一个更新
In [81]:
df = pd.DataFrame({'col1' : ["a","b",3,'?'],
'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 ? 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 ? ü á ?*ü*á
My code still works with Spanish characters
我的代码仍然适用于西班牙语字符
回答by fixxxer
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 ?*ü*á
dtype: object
回答by Anish Shah
You can use dropna()
您可以使用 dropna()
df['col4'] = df.apply(lambda row: '*'.join(row.dropna()), axis=1)
UPDATE:
更新：
Since, you need to convert numbers and special chars too, you can use astype(unicode)
因为，你也需要转换数字和特殊字符，你可以使用 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
回答by Zah
df.apply(lambda row: '*'.join(row.dropna()), axis=1)
回答by Julien Spronck
for row in xrange(len(df)):
s = '*'.join(df.ix[row].dropna().tolist())
print s
