如何检查字符串中的第n个字符，然后在新列中更新Python [英] How to check the n-th character in string , then update in new column Python

问题描述

目标:如果字符串匹配条件中的字符的第8个(或第n个)，则然后在新列中更新

Target : if the 8th (or n of ) of character in string match condition, then update in new column

按单词输入一个字符串:

# if i want to check the 3rd character
IN[0]:  s = "apple"
        s[2]
OUT[0]: 'p'

代码:

tt = pd.DataFrame({"CC":["T020203J71500","Y020203K71500","T020407JLX100","P020403JLX100"])

tt["NAME"] = pd.np.where(tt["CC"][7].str.contains("J"),"JANICE",
               pd.np.where(tt["CC"][7].str.contains("K"),"KELVIN",
               pd.np.where(tt["CC"][7].str.contains("X"),"SPECIAL","NONE")))

问题: 显然[7]不是python练习

Problem : Apparently [7] is not a python practice

在R data.table中:

tt[grepl("J",str_sub(CC,8,8)),
      "NAME":="JANICE"]
tt[grepl("K",str_sub(CC,8,8)),
      "NAME":="KELVIN"] # .... can achieve by doing like this 

如何在Python中执行此操作?

How can i do this in Python ?

推荐答案

您可以使用 np.select 此处，因为我们有多个条件:

You can use np.select here since we have multiple conditions:

conditions = [
    tt['CC'].str[7].eq('J'),
    tt['CC'].str[7].eq('K'),
    tt['CC'].str[7].eq('X')
]

choices = ['JANICE', 'KELVIN', 'SPECIAL']

tt['NAME'] = np.select(conditions, choices, default='NONE')

输出

              CC    NAME
0  T020203J71500  JANICE
1  Y020203K71500  KELVIN
2  T020407JLX100  JANICE
3  P020403JLX100  JANICE

这篇关于如何检查字符串中的第n个字符，然后在新列中更新Python的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！