数据似乎是XML格式的（尽管〈Order No. 10〉.使其无效;我已经将其与交换。您可以遍历每个值并检查它们是否以#CB开始。或者，您可以执行如下所示的正则表达式搜索。但是，如果您确信数据始终遵循XML格式，则前一种方法更合适。

import re

text = ...

# Extract between [#CB: and ]
cb_data = re.findall(r"\[#CB:(.*?)\]", text)
print(cb_data) # prints out CB

# Extract between [CR: and ]
cr_data = re.findall(r"\[CR:(.*?)\]", text)
print(cr_data) # prints out CR

如果您的数据始终是XML格式：

import xml.etree.ElementTree as ET

xml_str1 = '<Order><Remark>Food was good</Remark><UserID>7890</UserID><Filter>[#CB:Customer happy with service]</Filter><Rating>Five</Rating></Order>'
xml_str2 = '<Order><UserID>7880</UserID><Filter>[#CB:Customer had a good time</Filter><Remark>Food was good</Remark><Additional>Service up to par</Additional><Remark>(#CB:Customer will return again]</Remark><End>Thats all</End></Order>'

root1 = ET.fromstring(xml_str1)
root2 = ET.fromstring(xml_str2)

cb_filter1 = root1.find('Filter').text
cb_data1 = cb_filter1[cb_filter1.find('#CB:')+4:cb_filter1.find(']')]

cb_filter2 = root2.find('Filter').text
cb_data2 = cb_filter2[cb_filter2.find('#CB:')+4:cb_filter2.find(']')]

print(cb_data1) # Customer happy with service
print(cb_data2) # Customer had a good time

在另一个答案的基础上，这就是你如何以你要求的格式得到它。

import re
import pandas as pd

p = re.compile(r"\[(.*?)\]")

s1 = "<Order No. 10>Food was good7890[#CB:Customer happy with service]Five"
s2 = " Five<Order No. 17>7880[#CB:Customer had a good time]Food was goodService up to par[#CB:Customer will return again]Thats all"
print(p.findall(s1))
print(p.findall(s2))

d = pd.DataFrame({'scenario':[1,2], 'order_id':['1234', '1235'], 'reviews':[s1,s2] })

def padList(l, length=4):
  l = l[:length]
  l += [''] * (length - len(l))
  return l

d[['review1','review2','review3','review4']] = d.apply(lambda row: padList(p.findall(row['reviews'])), axis=1, result_type='expand')

d