从字典列表中的字符串编号中删除逗号（，）答案

【问题标题】：Remove comma(,) from string number in dictionary list从字典列表中的字符串编号中删除逗号（，）
【发布时间】：2020-08-21 01:12:08
【问题描述】：

我有一个字典列表:: 示例数据:: 像这样我有 n 个数据。

datas = [{"_id":"1234as", "Total students":"123,321", "TotalPresent":"321,345"}, 
    {"_id":"1234asas","TotalStudents":"343,431","TotalPresent":"541,656"}]

我试过了

for data in datas:
    for i in data.values():
        re.sub('[^A-Za-z0-9]+', '', i)
    datas.append(i)

我只想从 TotalStudents 和 TotalPresent 中删除逗号 (,) 并替换 datas 中的值。

编辑 1

在我的字典列表中，我的值也为::

datas = [{"_id":"1234as","Totalstudents":"123,321","TotalPresent":"321,345"},      
        {"_id":"1234asas","TotalStudents":"343,431","TotalPresent":"541,656"},
         {"_id":"9934 asas","TotalStudents":"NA","TotalPresent":""}]

这里，TotalStudents 键的值为“NA”，TotalPresent 为“”。有没有办法将出现的“NA”或“”替换为“0”。

【问题讨论】：

标签： python list dictionary replace

【解决方案1】：

您可以遍历字典中的键值对。并在删除逗号后替换该键的值。

import re

datas = [{"_id": "1234as", "Total Students": "123,321", "TotalPresent": "321,345"},
         {"_id": "1234asas", "TotalStudents": "343,431", "TotalPresent": "541,656"}]


for data in datas:
    for key, value in data.items():
        print(key, value)
        value = re.sub('[^A-Za-z0-9]+', '', value)
        data[key] = value

print(datas)

结果

_id 1234as
Total Students 123,321
TotalPresent 321,345
_id 1234asas
TotalStudents 343,431
TotalPresent 541,656

[{'_id': '1234as', 'Total Students': '123321', 'TotalPresent': '321345'}, 
 {'_id': '1234asas', 'TotalStudents': '343431', 'TotalPresent': '541656'}]

这是一种使您的代码正常工作的方法，因此总是替换所有值。如有必要，您需要添加自己的检查以使其更智能。

编辑

为了捕捉 "NA" 和 "" 值，我添加了一些 if 语句。它很简单，并且与您自己的代码保持接近。

import re

datas = [{"_id":"1234as","TotalStudents":"123,321","TotalPresent":"321,345"},
        {"_id":"1234asas","TotalStudents":"343,431","TotalPresent":"541,656"},
         {"_id":"9934 asas","TotalStudents":"NA","TotalPresent":""}]


for data in datas:
    print(data)
    for key, value in data.items():

        if key == "TotalStudents":
            if value == "NA":
                value =  "0"
            else:
                value = re.sub('[^A-Za-z0-9]+', '', value)
        elif key == "TotalPresent":
            if not value:
                value = "0"
            else:
                value = re.sub('[^A-Za-z0-9]+', '', value)

        data[key] = value

print()
for data in datas:
    print(data)

结果

{'_id': '1234as', 'TotalStudents': '123321', 'TotalPresent': '321345'}
{'_id': '1234asas', 'TotalStudents': '343431', 'TotalPresent': '541656'}
{'_id': '9934 asas', 'TotalStudents': '0', 'TotalPresent': '0'}

为了使代码更高效，您可以将新值直接放入数据中。在这种情况下，您不再将“_id”替换为它自己的值。

import re

datas = [{"_id":"1234as","TotalStudents":"123,321","TotalPresent":"321,345"},
        {"_id":"1234asas","TotalStudents":"343,431","TotalPresent":"541,656"},
         {"_id":"9934 asas","TotalStudents":"NA","TotalPresent":""}]


for data in datas:
    print(data)
    for key, value in data.items():

        if key == "TotalStudents":
            if value == "NA":
                data[key] = "0"
            else:
                data[key] = re.sub('[^A-Za-z0-9]+', '', value)
        elif key == "TotalPresent":
            if not value:
                data[key] = "0"
            else:
                data[key] = re.sub('[^A-Za-z0-9]+', '', value)

print()
for data in datas:
    print(data)

【讨论】：

我发现您的回答对提出的问题很有帮助，并将接受此作为答案。但在我关闭此线程之前，您能否查看已编辑问题中提到的案例
我添加了一个简单的调整和一个额外的步骤，用于一些简单的效率提升。我希望它有所帮助。请注意，您的数据中有错字：“Totalstudents”。
将 0 更改为“0”以将字符串放回数据中

【解决方案2】：

re.sub 不能在原地工作 - 它确实返回更改后的str。更一般地说，strs 是处理它们的不可变函数，它们不能就地工作。使用re.sub 的解决方案可能如下所示：

import re
datas = [{"_id":"1234as","Total Students":"123,321","TotalPresent":"321,345"},
{"_id":"1234asas","TotalStudents":"343,431","TotalPresent":"541,656"}]
cleandatas = []
for data in datas:
    cleandatas.append({k:re.sub('[^A-Za-z0-9]+', '', v) for k,v in data.items()})
print(cleandatas)

输出：

[{'_id': '1234as', 'Total Students': '123321', 'TotalPresent': '321345'}, {'_id': '1234asas', 'TotalStudents': '343431', 'TotalPresent': '541656'}]

我使用dict-comprehension 创建了新的清理dicts

【讨论】：

【解决方案3】：

如果要替换特定键的值，请确保键相同，因为示例中的第一个字典具有 Total Students，而第二个字典具有 TotalStudents。

试试这个：

datas = [{"_id": "1234as", "Total Students": "123,321", "TotalPresent": "321,345"},
         {"_id": "1234asas", "Total Students": "343,431", "TotalPresent": "541,656"}]

for d in datas:
    d["Total Students"] = d["Total Students"].replace(",", "")
    d["TotalPresent"] = d["TotalPresent"].replace(",", "")

print(datas)
# output: [{'_id': '1234as', 'Total Students': '123321', 'TotalPresent': '321345'}, {'_id': '1234asas', 'Total Students': '343431', 'TotalPresent': '541656'}]

如果您想替换所有键中的逗号，您可以尝试（但请记住，在这种情况下，您的 dict 的所有值都必须是字符串）：

datas = [{"_id": "1234as", "Total Students": "123,321", "TotalPresent": "321,345"},
         {"_id": "1234asas", "Total Students": "343,431", "TotalPresent": "541,656"}]

for d in datas:
    for k in d:
        d[k] = d[k].replace(",", "")

【讨论】：