替换符合条件的所有单元格的内容

Question

当列名未知时，如何循环遍历整个数据框以删除包含特定字符串的单元格中的数据？

以下是我目前掌握的情况：

for (i in colnames(df)){
   df2 = df[~df[i].str.contains('found')]

我的数据：

  Getting links from: https://www.bar.com/ Getting links from: https://www.boo.com/ Getting links from: https://www.foo.com/
0           ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/
1             ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/
2         ├─BROKEN─ http://www.broken.com/     2 links found. 0 excluded. 0 broken.         ├─BROKEN─ http://www.broken.com/
3                                      NaN                                      NaN            ├───OK─── http://www.set.com/
4                                      NaN                                      NaN            ├───OK─── http://www.one.com/

如果一个单元格包含一个字符串，例如“找到”，我如何删除它的全部内容？我想删除单元格中的所有内容（包括字符串之前和之后。）

Answer 1

您可以用df.replace（{'test'：np.nan}）替换nan，这应该替换所有实例

Answer 2

由于您正在寻找一个字符串/值来检查并对整个DataFrame采取行动，因此DataFrame.replace方法符合这里的警告。.

数据帧示例：

>>> df
      a
0  foo1
1  foo2
2   bar
3   bar
4   bar

将条替换为空白，或者如果需要，可以将其替换为NaN值：

>>> df.replace("bar", "", regex=True)
      a
0  foo1
1  foo2
2
3
4

或者用NaN

>>> df.replace("bar", np.nan, regex=True)
 # df.replace("bar", np.nan, regex=True, inplace=True)
      a
0  foo1
1  foo2
2   NaN
3   NaN
4   NaN

如果要替换到实际数据帧中，您可以选择使用inplace=True：

模拟给出的示例：

>>> df
                                                                                                                         col1
0  Getting links from: https://www.bar.com/ Getting links from: https://www.boo.com/ Getting links from: https://www.foo.com/
1            ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/
2              ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/
3          ├─BROKEN─ http://www.broken.com/     2 links found. 0 excluded. 0 broken.         ├─BROKEN─ http://www.broken.com/
4                                       NaN                                      NaN            ├───OK─── http://www.set.com/
5                                       NaN                                      NaN            ├───OK─── http://www.one.com/

str的结果包含：

>>> df[~df["col1"].str.contains("found")]
                                                                                                                         col1
0  Getting links from: https://www.bar.com/ Getting links from: https://www.boo.com/ Getting links from: https://www.foo.com/
1            ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/           ├───OK─── http://www.this.com/
2              ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/             ├───OK─── http://www.is.com/
4                                       NaN                                      NaN            ├───OK─── http://www.set.com/
5                                       NaN                                      NaN            ├───OK─── http://www.one.com/

或者如上所述，确保值为字符串类型以应用该操作

>>> df[~df["col1"].astype(str).str.contains("found")]

Answer 3

你可以在这里使用appymap：

import re    

# just make sure all values are string type
df = df.astype(str)    

# remove the text which has word found
df.applymap(lambda x: re.sub('.*found.*','',x ))

                                          0
0  Getting links from: https://www.boo.com/
1            ├───OK─── http://www.this.com/
2              ├───OK─── http://www.is.com/
3                                          
4                                       nan
5                                       nan