Python 将整数列表转换为字符串

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时间:2020-08-19 01:14:50  来源:igfitidea点击:

Convert a list of integers to string

pythonstringlistint

提问by Mira Mira

I want to convert my list of integers into a string. Here is how I create the list of integers:

我想将我的整数列表转换为字符串。这是我创建整数列表的方法:

new = [0] * 6
for i in range(6):
    new[i] = random.randint(0,10)

Like this:

像这样:

new == [1,2,3,4,5,6]
output == '123456'

采纳答案by jgritty

There's definitely a slicker way to do this, but here's a very straight forward way:

肯定有一种更巧妙的方法可以做到这一点,但这里有一个非常直接的方法:

mystring = ""

for digit in new:
    mystring += str(digit)

回答by tynn

With Convert a list of characters into a stringyou can just do

使用Convert a list of characters into a string你可以做

''.join(map(str,new))

回答by jloup

Coming a bit late and somehow extending the question, but you could leverage the arraymodule and use:

来得有点晚,并以某种方式扩展了问题,但您可以利用该array模块并使用:

from array import array

array('B', new).tobytes()

b'\n\t\x05\x00\x06\x05'

In practice, it creates an array of 1-byte wide integers (argument 'B') from your list of integers. The array is then converted to a string as a binary data structure, so the output won't look as you expect (you can fix this point with decode()). Yet, it should be one of the fastest integer-to-string conversion methods and it should save some memory. See also documentation and related questions:

实际上,它会'B'从您的整数列表中创建一个 1 字节宽的整数数组(参数)。然后将数组转换为字符串作为二进制数据结构,因此输出不会像您预期的那样(您可以使用 修复这一点decode())。然而,它应该是最快的整数到字符串转换方法之一,它应该可以节省一些内存。另请参阅文档和相关问题:

https://www.python.org/doc/essays/list2str/

https://www.python.org/doc/essays/list2str/

https://docs.python.org/3/library/array.html#module-array

https://docs.python.org/3/library/array.html#module-array

Converting integer to string in Python?

在 Python 中将整数转换为字符串?

回答by Cherry

You can loop through the integers in the list while converting to string type and appending to "string" variable.

您可以循环遍历列表中的整数,同时转换为字符串类型并附加到“字符串”变量。

for int in list:
    string += str(int)

回答by Satyam Singh

two simple ways of doing this

两种简单的方法来做到这一点

"".join(map(str, A))
"".join([str(a) for a in A])