兩個列表之間的模式匹配-Python

Question

我在目錄中有一堆zip文件，如果其中一個缺少，我想得到通知。

示例代碼：

a = ['pattern1.zip', 'pattern2.zip', 'pattern3.zip']
b = []
for root,dirs,files in os.walk(some_path):
    for i in files:
        if i.endswith(('pattern1.zip', 'pattern2.zip', 'pattern3.zip')):
            b.append(i)

輸出：b = ['test-pattern1.zip'，'test-pattern2.zip'，'test-pattern3.zip']

想要將'b'的內容與'a'匹配，並檢查是否缺少任何zip文件

Answer 1

我會采取另一種方法：

patterns = {'pattern1.zip', 'pattern2.zip', 'pattern3.zip'}
for root, dirs, files in os.walk(some_path):
    for f in files:
        for pattern in patterns:
            if f.endswith(pattern):
                patterns.remove(pattern)
                break

print('Missing patterns:', patterns)

Answer 2

您可以將列表轉換為集合並采用它們的區別1 ：

files_that_should_be_present = ['pattern1.zip', 'pattern2.zip', 'pattern3.zip']
files_that_are_present = ['pattern1.zip', 'pattern2.zip']

files_missing = list(set(files_that_should_be_present) - set(files_that_are_present))
print(files_missing)

輸出： ['pattern3.zip']