Question

我在Python中有一个列表，其中包含每个用户配置文件发布的推文数量。

例如：

list = ['142', '1567', '153K', '32', '10', '50']

如您所见，当数字非常高（即153567）而不是实际数字时，会存储153K。

使用K替换列表中每个元素中的000字符（如果存在）的最简单方法是什么？

Answer 1

for (myVectorIterator = vector.begin(); myVectorIterator != vector.end(); myVectorIterator++)

在您的代码中，153K不是任何类型的有效文字

我将此列表解释为字符串列表。

Answer 2

使用替换功能

[i.replace('K', '000') for i in list]

然后你可以使用int

将它转换为list o integer

[int(i.replace('K', '000')) for i in list]

在这种情况下，此列表将返回：

[142, 1567, 153000, 32, 10, 50]

Answer 3

只需将list comprehension与replace一起使用即可。单线代码：

[ele.replace('K','000') for ele in l]       #l is the list

所以我们得到

IN : l = ['142', '1567', '153K', '32', '10', '50']
OUT : ['142', '1567', '153000', '32', '10', '50']

Answer 4

>> list = ['142', '1567', '153K', '32', '10', '50']
>> listNoK = [item.replace('K','000') for item in list]
>> listNoK

['142', '1567', '153000', '32', '10', '50']

或者获取数字：

>>> listNoK = [int(item.replace('K','000')) for item in list]
>>> listNoK
[142, 1567, 153000, 32, 10, 50]

Answer 5

解决此任务的另一种方法如下：

list = ['142', '1567', '153K', '32', '10', '50']

for index, item in enumerate(list):
    if 'K' in item:
        list[index] = item.replace('K', '000')

Answer 6

list = [142, 1567, '153K', 32, 10, 50]

comp_list = [x.replace('K', '000') if type(x) is str else x for x in list]

Answer 7

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输出：

data = ["142", "1567", "153K", "32", "10", "50"]
for i in range(len(data)):
    data[i] = data[i].replace('K','000')
print(data)

如何替换列表的每个元素中的特定字符

7 个答案: