Question

我有以下pandas数据框：

import pandas as pd
df = pd.DataFrame({ 'gene':["1 // foo // blabla",
                                   "2 // bar // lalala",
                                   "3 // qux // trilil",
                                   "4 // woz // hohoho"], 'cell1':[5,9,1,7], 'cell2':[12,90,13,87]})
df = source_df[["gene","cell1","cell2"]]

看起来像这样：

                 gene  cell1  cell2
0  1 // foo // blabla      5     12
1  2 // bar // lalala      9     90
2  3 // qux // trilil      1     13
3  4 // woz // hohoho      7     87

我想得到的是：

   gene    cell1  cell2
0   foo       5     12
1   bar       9     90
2   qux       1     13
3   woz       7     87

即按//选择拆分字符串的第二个元素作为分隔符。

我能做的最好的就是：

df["gene"] = df["gene"].str.split(" // ")
df
Out[17]:
               gene  cell1  cell2
0  [1, foo, blabla]      5     12
1  [2, bar, lalala]      9     90
2  [3, qux, trilil]      1     13
3  [4, woz, hohoho]      7     87

这样做的正确方法是什么？

Answer 1

使用矢量化str.split这比在大型数据集上使用apply要快得多：

In [13]:
df['gene'] = df['gene'].str.split('//').str[1]
df

Out[13]:
   cell1  cell2   gene
0      5     12   foo 
1      9     90   bar 
2      1     13   qux 
3      7     87   woz

Answer 2

您可以使用regex并按strip删除第一个和最后一个空格：

df["gene"] = df["gene"].str.extract(r"\/\/([a-z ]+)\/\/")
df["gene"] = df["gene"].str.strip()

print df
  gene  cell1  cell2
0  foo      5     12
1  bar      9     90
2  qux      1     13
3  woz      7     87

\/\/([a-z ]+)\/\/表示：

    \/ matches the character / literally
    \/ matches the character / literally
    1st Capturing group ([a-z ]+)
        [a-z ]+ match a single character present in the list below
            Quantifier: + Between one and unlimited times, as many times as possible,
            giving back as needed [greedy]
            a-z a single character in the range between a and z (case sensitive)
             the literal character  
    \/ matches the character / literally
    \/ matches the character / literally

没有条带的正则表达式：

df["gene"] = df["gene"].str.extract(r"\/\/\s*([a-z ]+)\s\/\/")

/\/\/\s*([a-z ]+)\s\/\//表示：

    \/ matches the character / literally
    \/ matches the character / literally
    \s* match any white space character [\r\n\t\f ]
        Quantifier: * Between zero and unlimited times, as many times as possible, 
        giving back as needed [greedy]
    1st Capturing group ([a-z ]+)
        [a-z ]+ match a single character present in the list below
            Quantifier: + Between one and unlimited times, as many times as possible, 
            giving back as needed [greedy]
            a-z a single character in the range between a and z (case sensitive)
            the literal character  
    \s match any white space character [\r\n\t\f ]
    \/ matches the character / literally
    \/ matches the character / literally

Answer 3

你非常接近，但是从结果分割中选择元素会更难以按照自己的方式进行。

以下是适用

的解决方案

>>> df['gene'] = df['gene'].apply(lambda s: s.split('//')[1])
>>> df

    gene  cell1  cell2
0   foo       5     12
1   bar       9     90
2   qux       1     13
3   woz       7     87

如何通过分隔符拆分pandas列并选择首选元素作为替换

3 个答案: