如何根據其中的字符對字符串列表進行排序？

Question

我有一個看起來像這樣的字符串列表：

['training_tech26.txt', 'training_tech41.txt', 'training_tech68.txt', 'training_tech84.txt', 'training_tech52.txt', 'training_sales17.txt', 'training_sales2.txt', 'training_tech47.txt', 'training_sales23.txt', 'training_sales3.txt', 'training_tech9.txt', 'training_tech12.txt']

我需要按正確的順序對這些文件進行排序，例如：

['training_tech1.txt', 'training_tech2.txt', 'training_sales3.txt', 'training_tech4.txt', 'training_tech5.txt']

我正在使用這些代碼來訪問我的文件夾中的所有文件，並將它們 append 放入一個列表中。 在文件夾本身中，它們的順序是正確的，所以我不知道為什么這個列表會亂七八糟。

tech_dir_path = "/path/to/folder/with/files"
res = []
tech_res = os.listdir(tech_dir_path)

Answer 1

有 python 用於字符串自然排序的第三方庫，稱為natsort ：

from natsort import natsorted
tech_res = ['training_tech26.txt', 'training_tech41.txt', 'training_tech68.txt', 'training_tech84.txt', 'training_tech52.txt', 'training_sales17.txt', 'training_sales2.txt', 'training_tech47.txt', 'training_sales23.txt', 'training_sales3.txt', 'training_tech9.txt', 'training_tech12.txt']

print(natsorted(tech_res, key=lambda y: y.lower()))

Output：

['training_sales2.txt', 'training_sales3.txt', 'training_sales17.txt', 'training_sales23.txt', 'training_tech9.txt', 'training_tech12.txt', 'training_tech26.txt', 'training_tech41.txt', 'training_tech47.txt', 'training_tech52.txt', 'training_tech68.txt', 'training_tech84.txt']

Answer 2

查看您的示例輸入，這是我的代碼：

input = ['training_tech26.txt', 'training_tech41.txt', 'training_tech68.txt', 'training_tech84.txt', 'training_tech52.txt', 'training_sales17.txt', 'training_sales2.txt', 'training_tech47.txt', 'training_sales23.txt', 'training_sales3.txt', 'training_tech9.txt', 'training_tech12.txt']

def sorter(input):
    idx = [int(''.join([j for j in i if j.isdigit()])) for i in input] # gets the numbers in int
    idx.sort() # sort numbers
    res = []
    for i in idx: # iterate over idx
        for j in input: # iterate over input
            if i == int(''.join([k for k in j if k.isdigit()])): # checks if i is same as the j input
                res.append(j)
    return res
print(sorter(input))

您還可以在這里查看更多方法。

Answer 3

您可以使用key參數進行sort以提取文件擴展名前面的數字。

import re

# TODO handle the possibility of a failed match
def get_number(s):
    n = re.search(r'(\d+).txt', s).group(1)
    return int(n)

然后，例如，

>>> files = ['training_tech26.txt', 'training_tech41.txt', 'training_tech68.txt', 'training_tech84.txt', 'training_tech52.txt', 'training_sales17.txt', 'training_sales2.txt', 'training_tech47.txt', 'training_sales23.txt', 'training_sales3.txt', 'training_tech9.txt', 'training_tech12.txt']

>>> for f in sorted(files, key=get_number):
...  print(f)
...
training_sales2.txt
training_sales3.txt
training_tech9.txt
training_tech12.txt
training_sales17.txt
training_sales23.txt
training_tech26.txt
training_tech41.txt
training_tech47.txt
training_tech52.txt
training_tech68.txt
training_tech84.txt

os.listdir返回一個按字母順序排序的列表，在這種情況下'12' < '9'因為1在9之前。