如何将由 mongo ObjectId 列表组成的字符串转换为仅包含 ids 的 Python 列表

Question

我有一个数据框，其中有一列包含 ObjectId 列表的字符串表示形式。即：

"[ObjectId('5d28938629fe749c7c12b6e3'), ObjectId('5caf4522a30528e3458b4579')]"

我想将它从字符串文字转换为仅包含 id 的 python 列表，例如：

['5d28938629fe749c7c12b6e3', '5caf4522a30528e3458b4579']

json.loads

和

ast.literal_eval

都失败，因为字符串包含

ObjectId

Answer 1

我分享这个正则表达式：https://regex101.com/r/m5rW2q/1

您可以单击代码生成器，例如：

import re

regex = r"ObjectId\('(\w+)'\)"

test_str = "[ObjectId('5d28938629fe749c7c12b6e3'), ObjectId('5caf4522a30528e3458b4579')]"

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

输出：

Match 1 was found at 1-37: ObjectId('5d28938629fe749c7c12b6e3')
Group 1 found at 11-35: 5d28938629fe749c7c12b6e3
Match 2 was found at 39-75: ObjectId('5caf4522a30528e3458b4579')
Group 1 found at 49-73: 5caf4522a30528e3458b4579

举个例子：

import re 
regex = r"ObjectId\('(\w+)'\)" 

test_str = "[ObjectId('5d28938629fe749c7c12b6e3'), ObjectId('5caf4522a30528e3458b4579')]" 

matches = re.finditer(regex, test_str, re.MULTILINE) 
[i.groups()[0] for i in matches]  

输出：

['5d28938629fe749c7c12b6e3', '5caf4522a30528e3458b4579']

有关正则表达式的所有信息都可以在这里找到：https://docs.python.org/3/library/re.html

Answer 2

好吧，你可以使用替换

a = "[ObjectId('5d28938629fe749c7c12b6e3'), ObjectId('5caf4522a30528e3458b4579')]"
a.replace('ObjectId(', '').replace(")","")
#Output:
"['5d28938629fe749c7c12b6e3', '5caf4522a30528e3458b4579']"

Answer 3

找到行；分割于 ';从列表中选择项目 1 和 3：

my_df.loc[my_df["my_column"].str.contains("ObjectId"),"my_column"].str.split("'")[0][1:4:2]

精确给出两个元素的列表：

['5d28938629fe749c7c12b6e3', '5caf4522a30528e3458b4579']

Answer 4

可以利用列表理解

list_of_str = [str(id) for id in list_of_ids]

这将为我们带来预期的结果。

['5d28938629fe749c7c12b6e3', '5caf4522a30528e3458b4579']