Question

我有一个Log df，在该df中，我有列Description。看起来像。

Description
Machine x : Turn off
Another action here
Another action here
Machine y : Turn off
Machine x : Turn on
Another action here

我只需要用“：”分割行

赞：

Description               Machine           Action
Machine x : Turn off      Machine x         Turn off
Another action here
Another action here
Machine y : Turn off      Machine y         Turn off
Machine x : Turn on       Machine x         Turn on
Another action here

我已经尝试过：

s = df["Description"].apply(lambda x:x.split(":"))
df["Action"] = s.apply(lambda x: x[1])
df["Machine"] = s.apply(lambda x: x[0])

还有带有“ startswith”的内容。

Answer 1

给出一个数据框

>>> df
            Description
0  Machine x : Turn off
1   Another action here
2   Another action here
3  Machine y : Turn off
4   Machine x : Turn on
5   Another action here

我会通过Series.str.split(splitter, expand=True)来解决这个问题。

>>> has_colon = df['Description'].str.contains(':')
>>> df[['Machine', 'Action']] = df.loc[has_colon, 'Description'].str.split('\s*:\s*', expand=True)
>>> df
            Description    Machine    Action
0  Machine x : Turn off  Machine x  Turn off
1   Another action here        NaN       NaN
2   Another action here        NaN       NaN
3  Machine y : Turn off  Machine y  Turn off
4   Machine x : Turn on  Machine x   Turn on
5   Another action here        NaN       NaN

如果您希望使用空字符串，则可以通过以下方式替换NaN单元格

>>> df.fillna('')
            Description    Machine    Action
0  Machine x : Turn off  Machine x  Turn off
1   Another action here                     
2   Another action here                     
3  Machine y : Turn off  Machine y  Turn off
4   Machine x : Turn on  Machine x   Turn on
5   Another action here

Answer 2

您可以将str.extract与合适的regex一起使用。这将找到:周围的所有值（还会去除冒号周围的空格）：

df[['Machine', 'Action']] = df.Description.str.extract('(.*) : (.*)',expand=True)

>>> df
            Description    Machine    Action
0  Machine x : Turn off  Machine x  Turn off
1   Another action here        NaN       NaN
2   Another action here        NaN       NaN
3  Machine y : Turn off  Machine y  Turn off
4   Machine x : Turn on  Machine x   Turn on
5   Another action here        NaN       NaN

# df[['Machine', 'Action']] = df.Description.str.extract('(.*) : (.*)',expand=True).fillna('')

Answer 3

只需将split与expand=True一起使用

df[['Machine', 'Action']] =df.Description.str.split(':',expand=True).dropna()
df
            Description     Machine     Action
0  Machine x : Turn off  Machine x    Turn off
1   Another action here         NaN        NaN
2   Another action here         NaN        NaN
3  Machine y : Turn off  Machine y    Turn off
4   Machine x : Turn on  Machine x     Turn on
5   Another action here         NaN        NaN

Answer 4

具有pd.Series.str.extract函数和特定的正则表达式模式（在:分隔符周围包含潜在的多个空格）：

In [491]: df
Out[491]: 
            Description
0  Machine x : Turn off
1   Another action here
2   Another action here
3  Machine y : Turn off
4   Machine x : Turn on
5   Another action here

In [492]: pd.concat([df, df.Description.str.extract('(?P<Machine>[^:]+)\s+:\s+(?P<Action>[^:]+)').fillna('')], axis=1)
Out[492]: 
            Description    Machine    Action
0  Machine x : Turn off  Machine x  Turn off
1   Another action here                     
2   Another action here                     
3  Machine y : Turn off  Machine y  Turn off
4   Machine x : Turn on  Machine x   Turn on
5   Another action here

Answer 5

StringMethods有用且方便，但通常效果不佳。

我建议使用默认构造函数和纯python字符串处理

df[['Machine', 'Action']] = pd.DataFrame([x.split(':') for x in df.Description]).dropna()

计时优于.str访问器选项。

df = pd.concat([df]*1000)
%timeit pd.DataFrame([x.split(':') for x in df.Description]).dropna()
4.47 ms ± 252 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df.Description.str.split(':',expand=True).dropna()
14.9 ms ± 323 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df.Description.str.extract('(.*) : (.*)',expand=True)
16.6 ms ± 393 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit pd.concat([df, df.Description.str.extract('(?P<Machine>[^:]+)\s+:\s+(?P<Action>[^:]+)').fillna('')], axis=1)
22.5 ms ± 448 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 6

我的主张是：

msk = df.Description.str.contains(':')
df[['Machine', 'Action']] = df.Description.str.split(':', 1, expand=True).where(msk, '')

首先创建一个掩码-行可以接收非空值。

然后仅对掩码为true的行执行实际替换。其他行（实际上是所有新列）收到一个空字符串。

如何用条件分割熊猫的列

6 个答案: