Python 比较两个文件夹,返回不同文件的完整路径
我有一个脚本,可以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Python 比较两个文件夹,返回不同文件的完整路径,python,comparison,list-comprehension,Python,Comparison,List Comprehension,我有一个脚本,可以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(
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。有人知道如何解决这个问题吗
这将是理想的结果:
['C:/Users/langma/Desktop/EDI/Downloadfolder/EDI_2020-05-18\\file5.xlsx']
你可以通过改变获得结果的方式来做到这一点
result=[list1[i]表示范围内的i(len(list1_short)),如果list1_short[i]不在list2_short]
您可以通过改变获得结果的方式来实现这一点
result=[list1[i]表示范围内的i(len(list1_short)),如果list1_short[i]不在list2_short]
实际上,你可以去掉列表2\u 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]
实际上,你可以去掉
list2\u 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]
您可能有兴趣了解
zip()
和enumerate()
函数……您可能有兴趣了解zip()
和enumerate()
函数。。。