打印路径列表中任何项目的出现次数

Question

我正在使用os.walk 来识别通用源目录 (SRC) 中包含my_list 中的任何字符串的路径：

SRC = '/User/dir_1/'

my_list = ["dog", "cat", "mouse", "bird"]

for dirpath, dirnames, filenames in os.walk(SRC):
    for folders in dirnames:
        for x in my_list:
            if x in folders:
                source_path = os.path.join(dirpath, folders)

假设print(source_path) 给出以下内容：

/User/dir_1/cat_test/
/User/dir_1/cat_test/bird_results/
/User/dir_1/dir_2/dog_test/
/User/dir_1/dir_2/dog_test/cat_results/
/User/dir_1/mouse_test/
/User/dir_1/mouse_test/mouse_results/
/User/dir_1/unknown_test/dog_results/
/User/dir_1/bird_files/
/User/dir_1/bird_files/bird_a_files/
/User/dir_1/bird_files/bird_b_files/

我的目标是 shutil.move 我的 source_path 的，但是因为，例如，移动 /User/dir_1/bird_files/ 然后尝试移动 /User/dir_1/bird_files/bird_a_files/ 将导致 FileNotFound 错误，我想过滤我的 source_path 以包括那些只有 1 次出现的任何my_list 中的字符串，这样我的 source_path 是：

/User/dir_1/cat_test/
/User/dir_1/dir_2/dog_test/
/User/dir_1/mouse_test/
/User/dir_1/unknown_test/dog_results/
/User/dir_1/bird_files/

我已经尝试过source_path.count(x) == 1，但它迭代了my_list，而不是计数any x in my_list，这样我的输出是（例如）：

/User/dir_1/dir_2/dog_test/cat_results/ count == 1 (for dog)
/User/dir_1/dir_2/dog_test/cat_results/ count == 1 (for cat)
/User/dir_1/dir_2/dog_test/cat_results/ count == 0 (for mouse)
/User/dir_1/dir_2/dog_test/cat_results/ count == 0 (for bird)

但我想看看（例如）：

/User/dir_1/dir_2/dog_test/cat_results/ count == 2 (for any x in my_list)

这将允许我用count != 1 过滤掉任何source_path

Answer 1

使用推导按计数过滤，然后对结果求和（True 被强制转换为 1）以获得“任何”行为。

paths = """/User/dir_1/cat_test/
/User/dir_1/cat_test/bird_results/
/User/dir_1/dir_2/dog_test/
/User/dir_1/dir_2/dog_test/cat_results/
/User/dir_1/mouse_test/
/User/dir_1/mouse_test/mouse_results/
/User/dir_1/unknown_test/dog_results/
/User/dir_1/bird_files/
/User/dir_1/bird_files/bird_a_files/
/User/dir_1/bird_files/bird_b_files/""".split()


my_list = ["dog", "cat", "mouse", "bird"]

out = []
for path in paths:
    if sum(True for term in my_list if path.count(term) == 1) == 1:
        out.append(path)

print(*out, sep='
')

输出

/User/dir_1/cat_test/
/User/dir_1/dir_2/dog_test/
/User/dir_1/mouse_test/
/User/dir_1/unknown_test/dog_results/
/User/dir_1/bird_files/

---

编辑：从评论中，os.walk 方法。

想法：从dirnames 参数中删除术语

备注：我用作过滤条件（见代码中的注释）的方法子字符串包含在字符串中这是很差的。在这种特殊情况下，更健壮的可能是d.startswith(c)。为了获得更大的灵活性，请使用正则表达式- 类似的解决方案。

import os


constraints = 'dog', 'cat', 'mouse', 'bird'

wdir = './User' # your reference directory
res = []
for path, dirs, _ in os.walk(wdir, topdown=True):
    # local to each directory's content
    counter = dict.fromkeys(constraints, False)
    dirs_to_skip = []
    
    # filter by constraint
    for c in constraints:
        for d in dirs:
            if c in d: # <-- filter condition!
                if not counter[c]: # 1st match
                    counter[c] = True
                    res.append(os.path.join(path, d))

                dirs_to_skip.append(d)
    
    # remove matched paths          
    for d in dirs_to_skip:
        dirs.remove(d)

print(*res, sep='
')