比较两个文件夹，返回不同文件的完整路径。

Question

我有一个脚本，它应该比较文件夹和子文件夹中的文件。新文件应稍后复制。这是我用来创建列表的函数。

def fullNames(source):
    matches = []

    for root, dirnames, filenames in os.walk(source):
        for filename in filenames:
            if filename.endswith('.xlsx'):
                matches.append(os.path.join(root, filename))
    return matches

此函数返回如下列表：

list1 =  ['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file1.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file2.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file3.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file4.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file5.xlsx']

list2 =  ['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17\\file1.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17\\file2.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17\\file3.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17\\file4.xlsx']

要比较这些文件，我必须比较每个文件的基本名称。

list1_short = [os.path.basename(file) for file in list1]
list2_short = [os.path.basename(file) for file in list2]

result = [item for item in list1_short if item not in list2_short]
result

Out[134]: ['file5.xlsx']

这是可行的，但我需要返回该文件的完整路径，而不是basename。有谁知道怎么解决这个问题吗？

这将是预期的结果：

['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file5.xlsx']

Answer 1

实际上，你可以摆脱list2_short：

list1 =  ['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18/file1.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18/file2.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18/file3.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18/file4.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18/file5.xlsx']

list2 =  ['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17/file1.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17/file2.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17/file3.xlsx',
          'C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-17/file4.xlsx']


existing_names = [os.path.basename(item) for item in list2]
missing_files = [item for item in list1 if os.path.basename(item) not in existing_names]

Answer 2

你可以通过改变你获得结果的方式来做到这一点，

result = [list1[i] for i in range(len(list1_short)) if list1_short[i] not in list2_short]

Answer 3

list1_short = [os.path.basename(file) for file in list1]
list2_short = [os.path.basename(file) for file in list2]
missing_files=[]
for i,item in enumerate(list1_short):
    if item not in list2_short:
        missing_files.append(list1[i])