问题:如何将一列分为两列?

我有一个带有一列的数据框,我想将其分为两列,其中一列标题为’ fips',另一列为'row'

我的数据框df如下所示:

          row
0    00000 UNITED STATES
1    01000 ALABAMA
2    01001 Autauga County, AL
3    01003 Baldwin County, AL
4    01005 Barbour County, AL

我不知道如何使用df.row.str[:]以达到分割行单元的目的。我可以df['fips'] = hello用来添加一个新列,并用填充它hello。有任何想法吗?

         fips       row
0    00000 UNITED STATES
1    01000 ALABAMA 
2    01001 Autauga County, AL
3    01003 Baldwin County, AL
4    01005 Barbour County, AL

I have a data frame with one (string) column and I’d like to split it into two (string) columns, with one column header as ‘fips' and the other 'row'

My dataframe df looks like this:

          row
0    00000 UNITED STATES
1    01000 ALABAMA
2    01001 Autauga County, AL
3    01003 Baldwin County, AL
4    01005 Barbour County, AL

I do not know how to use df.row.str[:] to achieve my goal of splitting the row cell. I can use df['fips'] = hello to add a new column and populate it with hello. Any ideas?

         fips       row
0    00000 UNITED STATES
1    01000 ALABAMA 
2    01001 Autauga County, AL
3    01003 Baldwin County, AL
4    01005 Barbour County, AL

回答 0

也许有更好的方法,但这是一种方法:

                            row
    0       00000 UNITED STATES
    1             01000 ALABAMA
    2  01001 Autauga County, AL
    3  01003 Baldwin County, AL
    4  01005 Barbour County, AL
df = pd.DataFrame(df.row.str.split(' ',1).tolist(),
                                 columns = ['flips','row'])
   flips                 row
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

There might be a better way, but this here’s one approach:

                            row
    0       00000 UNITED STATES
    1             01000 ALABAMA
    2  01001 Autauga County, AL
    3  01003 Baldwin County, AL
    4  01005 Barbour County, AL
df = pd.DataFrame(df.row.str.split(' ',1).tolist(),
                                 columns = ['flips','row'])
   flips                 row
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

回答 1

TL; DR版本:

对于以下简单情况:

  • 我有一个带有定界符的文本列,我想要两列

最简单的解决方案是:

df['A'], df['B'] = df['AB'].str.split(' ', 1).str

或者,您可以使用以下方法自动为拆分的每个条目创建一个带有一列的DataFrame:

df['AB'].str.split(' ', 1, expand=True)

expand=True如果字符串的分割数不一致,并且要None替换缺失的值,则必须使用。

请注意,无论哪种情况,该.tolist()方法都是不必要的。都不是zip()

详细地:

安迪·海登(Andy Hayden)的解决方案最能证明该str.extract()方法的强大功能。

但是对于在已知分隔符上的简单拆分(例如,用破折号拆分或通过空格拆分),该.str.split()方法就足够了1。它对字符串的一列(系列)进行操作,并返回列表的一列(系列):

>>> import pandas as pd
>>> df = pd.DataFrame({'AB': ['A1-B1', 'A2-B2']})
>>> df

      AB
0  A1-B1
1  A2-B2
>>> df['AB_split'] = df['AB'].str.split('-')
>>> df

      AB  AB_split
0  A1-B1  [A1, B1]
1  A2-B2  [A2, B2]

1:如果不确定.str.split()do 的前两个参数是什么,我建议使用该方法纯Python版本的文档。

但是你如何去做:

  • 包含两个元素的列表的列

至:

  • 两列,每列包含列表的相应元素?

好吧,我们需要仔细查看.str列的属性。

这是一个神奇的对象,用于收集将列中的每个元素视为字符串的方法,然后在每个元素中尽可能有效地应用相应的方法:

>>> upper_lower_df = pd.DataFrame({"U": ["A", "B", "C"]})
>>> upper_lower_df

   U
0  A
1  B
2  C
>>> upper_lower_df["L"] = upper_lower_df["U"].str.lower()
>>> upper_lower_df

   U  L
0  A  a
1  B  b
2  C  c

但是它也有一个“索引”接口,用于通过其索引获取字符串的每个元素:

>>> df['AB'].str[0]

0    A
1    A
Name: AB, dtype: object

>>> df['AB'].str[1]

0    1
1    2
Name: AB, dtype: object

当然,.str只要可以对其建立索引,则此索引接口并不真正在乎它所索引的每个元素是否实际上是字符串,因此:

>>> df['AB'].str.split('-', 1).str[0]

0    A1
1    A2
Name: AB, dtype: object

>>> df['AB'].str.split('-', 1).str[1]

0    B1
1    B2
Name: AB, dtype: object

然后,只需利用Python元组对可迭代对象进行拆包即可

>>> df['A'], df['B'] = df['AB'].str.split('-', 1).str
>>> df

      AB  AB_split   A   B
0  A1-B1  [A1, B1]  A1  B1
1  A2-B2  [A2, B2]  A2  B2

当然,从拆分一列字符串中获取一个DataFrame非常有用,以至于该.str.split()方法可以通过expand=True参数为您做到这一点:

>>> df['AB'].str.split('-', 1, expand=True)

    0   1
0  A1  B1
1  A2  B2

因此,完成我们想要的工作的另一种方法是:

>>> df = df[['AB']]
>>> df

      AB
0  A1-B1
1  A2-B2

>>> df.join(df['AB'].str.split('-', 1, expand=True).rename(columns={0:'A', 1:'B'}))

      AB   A   B
0  A1-B1  A1  B1
1  A2-B2  A2  B2

expand=True版本虽然较长,但与元组拆包方法相比具有明显的优势。元组解压缩不能很好地处理不同长度的拆分:

>>> df = pd.DataFrame({'AB': ['A1-B1', 'A2-B2', 'A3-B3-C3']})
>>> df
         AB
0     A1-B1
1     A2-B2
2  A3-B3-C3
>>> df['A'], df['B'], df['C'] = df['AB'].str.split('-')
Traceback (most recent call last):
  [...]    
ValueError: Length of values does not match length of index
>>> 

但是expand=True通过放置None没有足够“拆分”的列来很好地处理它:

>>> df.join(
...     df['AB'].str.split('-', expand=True).rename(
...         columns={0:'A', 1:'B', 2:'C'}
...     )
... )
         AB   A   B     C
0     A1-B1  A1  B1  None
1     A2-B2  A2  B2  None
2  A3-B3-C3  A3  B3    C3

TL;DR version:

For the simple case of:

  • I have a text column with a delimiter and I want two columns

The simplest solution is:

df[['A', 'B']] = df['AB'].str.split(' ', 1, expand=True)

You must use expand=True if your strings have a non-uniform number of splits and you want None to replace the missing values.

Notice how, in either case, the .tolist() method is not necessary. Neither is zip().

In detail:

Andy Hayden’s solution is most excellent in demonstrating the power of the str.extract() method.

But for a simple split over a known separator (like, splitting by dashes, or splitting by whitespace), the .str.split() method is enough1. It operates on a column (Series) of strings, and returns a column (Series) of lists:

>>> import pandas as pd
>>> df = pd.DataFrame({'AB': ['A1-B1', 'A2-B2']})
>>> df

      AB
0  A1-B1
1  A2-B2
>>> df['AB_split'] = df['AB'].str.split('-')
>>> df

      AB  AB_split
0  A1-B1  [A1, B1]
1  A2-B2  [A2, B2]

1: If you’re unsure what the first two parameters of .str.split() do, I recommend the docs for the plain Python version of the method.

But how do you go from:

  • a column containing two-element lists

to:

  • two columns, each containing the respective element of the lists?

Well, we need to take a closer look at the .str attribute of a column.

It’s a magical object that is used to collect methods that treat each element in a column as a string, and then apply the respective method in each element as efficient as possible:

>>> upper_lower_df = pd.DataFrame({"U": ["A", "B", "C"]})
>>> upper_lower_df

   U
0  A
1  B
2  C
>>> upper_lower_df["L"] = upper_lower_df["U"].str.lower()
>>> upper_lower_df

   U  L
0  A  a
1  B  b
2  C  c

But it also has an “indexing” interface for getting each element of a string by its index:

>>> df['AB'].str[0]

0    A
1    A
Name: AB, dtype: object

>>> df['AB'].str[1]

0    1
1    2
Name: AB, dtype: object

Of course, this indexing interface of .str doesn’t really care if each element it’s indexing is actually a string, as long as it can be indexed, so:

>>> df['AB'].str.split('-', 1).str[0]

0    A1
1    A2
Name: AB, dtype: object

>>> df['AB'].str.split('-', 1).str[1]

0    B1
1    B2
Name: AB, dtype: object

Then, it’s a simple matter of taking advantage of the Python tuple unpacking of iterables to do

>>> df['A'], df['B'] = df['AB'].str.split('-', 1).str
>>> df

      AB  AB_split   A   B
0  A1-B1  [A1, B1]  A1  B1
1  A2-B2  [A2, B2]  A2  B2

Of course, getting a DataFrame out of splitting a column of strings is so useful that the .str.split() method can do it for you with the expand=True parameter:

>>> df['AB'].str.split('-', 1, expand=True)

    0   1
0  A1  B1
1  A2  B2

So, another way of accomplishing what we wanted is to do:

>>> df = df[['AB']]
>>> df

      AB
0  A1-B1
1  A2-B2

>>> df.join(df['AB'].str.split('-', 1, expand=True).rename(columns={0:'A', 1:'B'}))

      AB   A   B
0  A1-B1  A1  B1
1  A2-B2  A2  B2

The expand=True version, although longer, has a distinct advantage over the tuple unpacking method. Tuple unpacking doesn’t deal well with splits of different lengths:

>>> df = pd.DataFrame({'AB': ['A1-B1', 'A2-B2', 'A3-B3-C3']})
>>> df
         AB
0     A1-B1
1     A2-B2
2  A3-B3-C3
>>> df['A'], df['B'], df['C'] = df['AB'].str.split('-')
Traceback (most recent call last):
  [...]    
ValueError: Length of values does not match length of index
>>> 

But expand=True handles it nicely by placing None in the columns for which there aren’t enough “splits”:

>>> df.join(
...     df['AB'].str.split('-', expand=True).rename(
...         columns={0:'A', 1:'B', 2:'C'}
...     )
... )
         AB   A   B     C
0     A1-B1  A1  B1  None
1     A2-B2  A2  B2  None
2  A3-B3-C3  A3  B3    C3

回答 2

您可以使用正则表达式模式整齐地提取不同部分:

In [11]: df.row.str.extract('(?P<fips>\d{5})((?P<state>[A-Z ]*$)|(?P<county>.*?), (?P<state_code>[A-Z]{2}$))')
Out[11]: 
    fips                    1           state           county state_code
0  00000        UNITED STATES   UNITED STATES              NaN        NaN
1  01000              ALABAMA         ALABAMA              NaN        NaN
2  01001   Autauga County, AL             NaN   Autauga County         AL
3  01003   Baldwin County, AL             NaN   Baldwin County         AL
4  01005   Barbour County, AL             NaN   Barbour County         AL

[5 rows x 5 columns]

要解释有点长的正则表达式:

(?P<fips>\d{5})
  • 匹配五个数字(\d)并命名"fips"

下一部分:

((?P<state>[A-Z ]*$)|(?P<county>.*?), (?P<state_code>[A-Z]{2}$))

|)做以下两件事之一:

(?P<state>[A-Z ]*$)
  • 匹配任意数量(*)的大写字母或空格([A-Z ]),并"state"在字符串($)末尾之前对其进行命名,

要么

(?P<county>.*?), (?P<state_code>[A-Z]{2}$))
  • .*然后匹配其他任何()
  • 然后用逗号和空格
  • 匹配state_code字符串($)末尾的两位数字。

在示例中:
请注意,前两行命中“州”(将NaN保留在county和state_code列中),而后三行命中县(即state_code)(将NaN保留在state列中)。

You can extract the different parts out quite neatly using a regex pattern:

In [11]: df.row.str.extract('(?P<fips>\d{5})((?P<state>[A-Z ]*$)|(?P<county>.*?), (?P<state_code>[A-Z]{2}$))')
Out[11]: 
    fips                    1           state           county state_code
0  00000        UNITED STATES   UNITED STATES              NaN        NaN
1  01000              ALABAMA         ALABAMA              NaN        NaN
2  01001   Autauga County, AL             NaN   Autauga County         AL
3  01003   Baldwin County, AL             NaN   Baldwin County         AL
4  01005   Barbour County, AL             NaN   Barbour County         AL

[5 rows x 5 columns]

To explain the somewhat long regex:

(?P<fips>\d{5})
  • Matches the five digits (\d) and names them "fips".

The next part:

((?P<state>[A-Z ]*$)|(?P<county>.*?), (?P<state_code>[A-Z]{2}$))

Does either (|) one of two things:

(?P<state>[A-Z ]*$)
  • Matches any number (*) of capital letters or spaces ([A-Z ]) and names this "state" before the end of the string ($),

or

(?P<county>.*?), (?P<state_code>[A-Z]{2}$))
  • matches anything else (.*) then
  • a comma and a space then
  • matches the two digit state_code before the end of the string ($).

In the example:
Note that the first two rows hit the “state” (leaving NaN in the county and state_code columns), whilst the last three hit the county, state_code (leaving NaN in the state column).


回答 3

df[['fips', 'row']] = df['row'].str.split(' ', n=1, expand=True)
df[['fips', 'row']] = df['row'].str.split(' ', n=1, expand=True)

回答 4

如果您不想创建新的数据框,或者您的数据框具有比仅要拆分的列更多的列,则可以:

df["flips"], df["row_name"] = zip(*df["row"].str.split().tolist())
del df["row"]  

If you don’t want to create a new dataframe, or if your dataframe has more columns than just the ones you want to split, you could:

df["flips"], df["row_name"] = zip(*df["row"].str.split().tolist())
del df["row"]  

回答 5

您可以使用空格(默认分隔符)和参数expand=True用于将DataFrame其分配给新列:

df = pd.DataFrame({'row': ['00000 UNITED STATES', '01000 ALABAMA', 
                           '01001 Autauga County, AL', '01003 Baldwin County, AL', 
                           '01005 Barbour County, AL']})
print (df)
                        row
0       00000 UNITED STATES
1             01000 ALABAMA
2  01001 Autauga County, AL
3  01003 Baldwin County, AL
4  01005 Barbour County, AL



df[['a','b']] = df['row'].str.split(n=1, expand=True)
print (df)
                        row      a                   b
0       00000 UNITED STATES  00000       UNITED STATES
1             01000 ALABAMA  01000             ALABAMA
2  01001 Autauga County, AL  01001  Autauga County, AL
3  01003 Baldwin County, AL  01003  Baldwin County, AL
4  01005 Barbour County, AL  01005  Barbour County, AL

修改是否需要删除原始列 DataFrame.pop

df[['a','b']] = df.pop('row').str.split(n=1, expand=True)
print (df)
       a                   b
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

什么是一样的:

df[['a','b']] = df['row'].str.split(n=1, expand=True)
df = df.drop('row', axis=1)
print (df)

       a                   b
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

如果出现错误:

#remove n=1 for split by all whitespaces
df[['a','b']] = df['row'].str.split(expand=True)

ValueError:列的长度必须与键的长度相同

您可以检查并返回4列DataFrame,不仅2:

print (df['row'].str.split(expand=True))
       0        1        2     3
0  00000   UNITED   STATES  None
1  01000  ALABAMA     None  None
2  01001  Autauga  County,    AL
3  01003  Baldwin  County,    AL
4  01005  Barbour  County,    AL

然后将解决方案添加到新DataFramejoin

df = pd.DataFrame({'row': ['00000 UNITED STATES', '01000 ALABAMA', 
                           '01001 Autauga County, AL', '01003 Baldwin County, AL', 
                           '01005 Barbour County, AL'],
                    'a':range(5)})
print (df)
   a                       row
0  0       00000 UNITED STATES
1  1             01000 ALABAMA
2  2  01001 Autauga County, AL
3  3  01003 Baldwin County, AL
4  4  01005 Barbour County, AL

df = df.join(df['row'].str.split(expand=True))
print (df)

   a                       row      0        1        2     3
0  0       00000 UNITED STATES  00000   UNITED   STATES  None
1  1             01000 ALABAMA  01000  ALABAMA     None  None
2  2  01001 Autauga County, AL  01001  Autauga  County,    AL
3  3  01003 Baldwin County, AL  01003  Baldwin  County,    AL
4  4  01005 Barbour County, AL  01005  Barbour  County,    AL

使用删除原始列(如果还有其他列):

df = df.join(df.pop('row').str.split(expand=True))
print (df)
   a      0        1        2     3
0  0  00000   UNITED   STATES  None
1  1  01000  ALABAMA     None  None
2  2  01001  Autauga  County,    AL
3  3  01003  Baldwin  County,    AL
4  4  01005  Barbour  County,    AL   

You can use by whitespace (default separator) and parameter expand=True for DataFrame with assign to new columns:

df = pd.DataFrame({'row': ['00000 UNITED STATES', '01000 ALABAMA', 
                           '01001 Autauga County, AL', '01003 Baldwin County, AL', 
                           '01005 Barbour County, AL']})
print (df)
                        row
0       00000 UNITED STATES
1             01000 ALABAMA
2  01001 Autauga County, AL
3  01003 Baldwin County, AL
4  01005 Barbour County, AL



df[['a','b']] = df['row'].str.split(n=1, expand=True)
print (df)
                        row      a                   b
0       00000 UNITED STATES  00000       UNITED STATES
1             01000 ALABAMA  01000             ALABAMA
2  01001 Autauga County, AL  01001  Autauga County, AL
3  01003 Baldwin County, AL  01003  Baldwin County, AL
4  01005 Barbour County, AL  01005  Barbour County, AL

Modification if need remove original column with DataFrame.pop

df[['a','b']] = df.pop('row').str.split(n=1, expand=True)
print (df)
       a                   b
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

What is same like:

df[['a','b']] = df['row'].str.split(n=1, expand=True)
df = df.drop('row', axis=1)
print (df)

       a                   b
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

If get error:

#remove n=1 for split by all whitespaces
df[['a','b']] = df['row'].str.split(expand=True)

ValueError: Columns must be same length as key

You can check and it return 4 column DataFrame, not only 2:

print (df['row'].str.split(expand=True))
       0        1        2     3
0  00000   UNITED   STATES  None
1  01000  ALABAMA     None  None
2  01001  Autauga  County,    AL
3  01003  Baldwin  County,    AL
4  01005  Barbour  County,    AL

Then solution is append new DataFrame by join:

df = pd.DataFrame({'row': ['00000 UNITED STATES', '01000 ALABAMA', 
                           '01001 Autauga County, AL', '01003 Baldwin County, AL', 
                           '01005 Barbour County, AL'],
                    'a':range(5)})
print (df)
   a                       row
0  0       00000 UNITED STATES
1  1             01000 ALABAMA
2  2  01001 Autauga County, AL
3  3  01003 Baldwin County, AL
4  4  01005 Barbour County, AL

df = df.join(df['row'].str.split(expand=True))
print (df)

   a                       row      0        1        2     3
0  0       00000 UNITED STATES  00000   UNITED   STATES  None
1  1             01000 ALABAMA  01000  ALABAMA     None  None
2  2  01001 Autauga County, AL  01001  Autauga  County,    AL
3  3  01003 Baldwin County, AL  01003  Baldwin  County,    AL
4  4  01005 Barbour County, AL  01005  Barbour  County,    AL

With remove original column (if there are also another columns):

df = df.join(df.pop('row').str.split(expand=True))
print (df)
   a      0        1        2     3
0  0  00000   UNITED   STATES  None
1  1  01000  ALABAMA     None  None
2  2  01001  Autauga  County,    AL
3  3  01003  Baldwin  County,    AL
4  4  01005  Barbour  County,    AL   

回答 6

如果要基于定界符将字符串分为两列以上,则可以省略“ maximum splits”参数。
您可以使用:

df['column_name'].str.split('/', expand=True)

这将自动创建与您的任何初始字符串中包含的最大字段数一样多的列。

If you want to split a string into more than two columns based on a delimiter you can omit the ‘maximum splits’ parameter.
You can use:

df['column_name'].str.split('/', expand=True)

This will automatically create as many columns as the maximum number of fields included in any of your initial strings.


回答 7

感到惊讶的是我还没有看到这个。如果您只需要两个分割,我强烈建议您。。。

Series.str.partition

partition 在分隔符上执行一次拆分,通常表现出色。

df['row'].str.partition(' ')[[0, 2]]

       0                   2
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

如果您需要重命名行,

df['row'].str.partition(' ')[[0, 2]].rename({0: 'fips', 2: 'row'}, axis=1)

    fips                 row
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

如果您需要将其恢复为原始版本,请使用joinconcat

df.join(df['row'].str.partition(' ')[[0, 2]])

pd.concat([df, df['row'].str.partition(' ')[[0, 2]]], axis=1)

                        row      0                   2
0       00000 UNITED STATES  00000       UNITED STATES
1             01000 ALABAMA  01000             ALABAMA
2  01001 Autauga County, AL  01001  Autauga County, AL
3  01003 Baldwin County, AL  01003  Baldwin County, AL
4  01005 Barbour County, AL  01005  Barbour County, AL

Surprised I haven’t seen this one yet. If you only need two splits, I highly recommend. . .

Series.str.partition

partition performs one split on the separator, and is generally quite performant.

df['row'].str.partition(' ')[[0, 2]]

       0                   2
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

If you need to rename the rows,

df['row'].str.partition(' ')[[0, 2]].rename({0: 'fips', 2: 'row'}, axis=1)

    fips                 row
0  00000       UNITED STATES
1  01000             ALABAMA
2  01001  Autauga County, AL
3  01003  Baldwin County, AL
4  01005  Barbour County, AL

If you need to join this back to the original, use join or concat:

df.join(df['row'].str.partition(' ')[[0, 2]])

pd.concat([df, df['row'].str.partition(' ')[[0, 2]]], axis=1)

                        row      0                   2
0       00000 UNITED STATES  00000       UNITED STATES
1             01000 ALABAMA  01000             ALABAMA
2  01001 Autauga County, AL  01001  Autauga County, AL
3  01003 Baldwin County, AL  01003  Baldwin County, AL
4  01005 Barbour County, AL  01005  Barbour County, AL

回答 8

我更喜欢导出相应的熊猫系列(即我需要的列),使用apply函数将列内容分为多个系列,然后加入生成的列到现有的数据帧。当然,应删除源列。

例如

 col1 = df["<col_name>"].apply(<function>)
 col2 = ...
 df = df.join(col1.to_frame(name="<name1>"))
 df = df.join(col2.toframe(name="<name2>"))
 df = df.drop(["<col_name>"], axis=1)

拆分两个单词的字符串函数应该是这样的:

lambda x: x.split(" ")[0] # for the first element
lambda x: x.split(" ")[-1] # for the last element

I prefer exporting the corresponding pandas series (i.e. the columns I need), using the apply function to split the column content into multiple series and then join the generated columns to the existing DataFrame. Of course, the source column should be removed.

e.g.

 col1 = df["<col_name>"].apply(<function>)
 col2 = ...
 df = df.join(col1.to_frame(name="<name1>"))
 df = df.join(col2.toframe(name="<name2>"))
 df = df.drop(["<col_name>"], axis=1)

To split two words strings function should be something like that:

lambda x: x.split(" ")[0] # for the first element
lambda x: x.split(" ")[-1] # for the last element

回答 9

我看到没有人使用过切片法,所以在这里我放了2美分。

df["<col_name>"].str.slice(stop=5)
df["<col_name>"].str.slice(start=6)

此方法将创建两个新列。

I saw that no one had used the slice method, so here I put my 2 cents here.

df["<col_name>"].str.slice(stop=5)
df["<col_name>"].str.slice(start=6)

This method will create two new columns.


回答 10

使用df.assign创建一个新的DF。参见http://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy

split = df_selected['name'].str.split(',', 1, expand=True)
df_split = df_selected.assign(first_name=split[0], last_name=split[1])
df_split.drop('name', 1, inplace=True)

Use df.assign to create a new df. See http://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy

split = df_selected['name'].str.split(',', 1, expand=True)
df_split = df_selected.assign(first_name=split[0], last_name=split[1])
df_split.drop('name', 1, inplace=True)

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