python - 如何对文件中的int进行排序

Question

我有这个代码：

import glob
import os

for file in glob.glob("./data/*.jpg"):
    lel=(os.path.basename(file))
    lel = (os.path.splitext(lel)[0])
    list1 = lel
    list1 = [int(x) for x in list1]
    list1.sort()
    print(list1)

它给出了这个输出：

[0]
[1]
[0, 1]
[1, 1]
[1, 2]
[1, 3]
[1, 4]
[1, 5]
[1, 6]
[1, 7]
[1, 8]
[1, 9]
[2]
[0, 2]
[1, 2]
[2, 2]
[2, 3]
[2, 4]
[2, 5]
[2, 6]
[2, 7]
[2, 8]
[2, 9]
[3]
[0, 3]
[1, 3]
[2, 3]
[3, 3]
[3, 4]
[4]
[5]
[6]
[7]
[8]
[9]

有没有办法让它输出这个？：

0
1
2
3
4
5
6
7
8
9
10
etc etc..

标签： pythonpython-os

Answer 1

此代码应该可以工作：

import glob
import os

list1 = []

for file in glob.glob("./data/*.jpg"): # iterate over filename
    list1.append(
                 os.path.splitext(
                                  os.path.basename(file)
                                 )[0]
    ) # strip away path and extension and append to the list1
print(*sorted(list1)) # print it out

您正在遍历文件名的字母，并将每个字母转换为 int。你甚至可以通过列表理解来做到这一点：

import glob
import os

list1 = [os.path.splitext(os.path.basename(filename))[0] for filename in glob.glob("./data/*.jpg")]
print(*sorted(list1))