获取列中每个条目中字符串的索引2-4。

huangapple
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英文:

How do I get the index 2-4 of a String in each entry of a column?

问题

我需要为数据框中的每个条目提取字符串的索引2-4。例如：

  1. A
  2. AB101CD
  3. AB101CD
  4. AB101CD

结果：

  1. A
  2. 101
  3. 101
  4. 101
英文:

I have a column in a dataframe and need for every entry just the index 2-4 of the string.
For example

  1. A
  2. AB101CD
  3. AB101CD
  4. AB101CD

result:

  1. A
  2. 101
  3. 101
  4. 101

答案1

得分: 0

你可以使用 df.column.apply：

  1. import pandas as pd    
  2. a = ['AB101CD','AB101CD','AB101CD']
  3. df = pd.DataFrame(a)
  4. df.columns = ['A']
  5. df.A.apply(lambda x:x[2:5])

这会得到以下结果：

  1. 0    101
  2. 1    101
  3. 2    101
  4. Name: A, dtype: object

你可以将其设置为列：

  1. df.A = df.A.apply(lambda x:x[2:5])
英文:

You could use df.column.apply

  1. import pandas as pd    
  2. a = ['AB101CD','AB101CD','AB101CD']
  3. df = pd.DataFrame(a)
  4. df.columns = ['A']
  5. df.A.apply(lambda x:x[2:5])

This gives:

  1. 0    101
  2. 1    101
  3. 2    101
  4. Name: A, dtype: object

You could set it as column:

  1.  df.A = df.A.apply(lambda x:x[2:5])

答案2

得分: 0

You can get a strings substring by using {string_variable}[{index1}:{index2}]

  1. str = "AB101CD"
  3. print(str[2:5])

output will be "101"

then you can apply this to your list or whatever with a for loop for example:

  1. mylist = ["A",
  2.           "AB101CD",
  3.           "AB101CD",
  4.           "AB101CD"]
  6. list2 = []
  8. for i in mylist:
  9.     #it wont return anything for the first element, "A", so you should define what do you want to do here for non-suitable elements
  10.     
  11.     #print
  12.     print(i[2:5])
  14.     #get into another list
  15.     list2.append(i[2:5])
英文:

You can get a strings substring by using {string_variable}[{index1}:{index2}]

  1. str = "AB101CD"
  3. print(str[2:5])

output will be "101"

then you can apply this to your list or whatever with a for loop for example:

  1. mylist = ["A",
  2.           "AB101CD",
  3.           "AB101CD",
  4.           "AB101CD"]
  6. list2 = []
  8. for i in mylist:
  9.     #it wont return anything for the first element, "A", so you should define what do you want to do here for non-suitable elements
  10.     
  11.     #print
  12.     print(i[2:5])
  14.     #get into another list
  15.     list2.append(i[2:5])

答案3

得分: 0

你可以使用广播操作来切割列

  1. df['A'] = df['A'].str[2:5]

输出

  1.      A
  2. 0  101
  3. 1  101
  4. 2  101
英文:

You can slice the column with broadcasting

  1. df['A'] = df['A'].str[2:5]

Output

  1.      A
  2. 0  101
  3. 1  101
  4. 2  101

huangapple
  • 本文由 发表于 2023年4月17日 15:41:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/76032745.html
  • dataframe
  • python
  • string
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