我有 1 个包含文件名的列表和 1 个包含过滤词的嵌套列表。过滤器列表有 3 个列表,每个列表的子列表长度不同。
如何迭代列表并使用 and
函数?由于 ['employer', 'finance']
和 ['employer', 'adress']
的差异,它需要检查列表中的所有值。
filter = [
['employer','finance'],
['manifest'],
['epmloyer','adress','home']
]
file_list = [
'01012017_employer_finance.txt',
'25102017_cargo_manifest.txt',
'12022014_employer_finance.txt',
'12022018_epmloyer_home_adress.txt',
'12032016_employer_home_adress.rtx'
]
"""search for financial file"""
if filter[0][0] in file_list[0] and filter[0][1] in file_list[0]:
print('Financial file found')
"""search for cargo manifest"""
if filter[1][0] in file_list[1]:
print('Cargo manifest found')
"""search for adress file"""
if filter[2][0] in file_list[2] and filter[2][1] in file_list[2] and filter[2][2] in file_list[2]:
print('Financial file found')
到目前为止,我设法获得了下面的代码。但是我如何处理不同长度的列表呢?以及使用变量,例如:filter[x][z]
代替 filter[1][0]
"""loop through the file_list"""
for file in file_list:
print("Identify file:", file)
#identify file in list with lists in it
if filter[0][0] in file and filter[0][1] in file:
print('***Financial file found')
好的,我使用了给出的代码。
file_list = [
'01012007-1_employer_finance.txt',
'25102013-2_cargo_manifest.txt',
'12022018-3_epmloyer_home_adress.txt',
'12022028-4_epmloyer_work_adress.txt',
'01012011-5_employer_finance.txt'
'01012007-12_employer_finance.txt',
'25102013-23_cargo_manifest.txt',
'12022018-34_epmloyer_home_adress.txt',
'12022028-45_epmloyer_work_adress.txt',
'01012011-56_employer_finance.txt'
]
"""Dictionary files"""
filters = {
'finance': ['employer','finance'],
'manifest': ['manifest'],
'address': ['epmloyer', 'adress', 'home'],
'addres': ['epmloyer', 'adress', 'work']
}
"""Tweede oplossing op stackoverflow"""
"""Loop through the nested list"""
def matches(filter, filename):
return all(x in filename for x in filter)
def get_filename(filter, files):
for f in files:
if matches(filter, f):
return f
for label, filter in filters.items():
file = get_filename(filter, file_list)
if file:
#print(f'Found {label} file: {file}')
pass
found_files = {label: get_filename(filters, file_list) for label, filters in filters.items()}
print(found_files)
结果是:
{'finance': '01012007-1_employer_finance.txt', 'manifest': '25102013-2_cargo_manifest.txt', 'address': '12022018-3_epmloyer_home_adress.txt', 'addres': '12022028-4_epmloyer_work_adress.txt'}
但是列表应该更大。
最佳答案
all
函数可用于根据文件名检查过滤器的元素:
def matches(filter, filename):
return all(x in filename for x in filter)
要查找与给定过滤器匹配的文件,您需要遍历文件列表并对每个项目应用 match
:
def get_filename(filter, files):
for f in files:
if matches(filter, f)
return f
这可以使用 next
函数以更短的方式表达:
def get_filename(filter, files):
return next((f for f in files if matches(filter, f)), None)
使用第二个参数调用 next
会使其返回 None
,而不是在没有匹配文件时引发错误。
现在您可以检查所有文件。我建议更进一步,使用字典来标记过滤器:
filters = {
'finance': ['employer','finance'],
'manifest': ['manifest'],
'address': ['epmloyer', 'adress', 'home'],
}
for label, filter in filters.items():
file = get_filename(filter, files)
if file:
print(f'Found {label} file: {file}')
您可以更进一步,为您找到的文件创建一个字典:
found_files = {label: get_filename(filter, files) for label, filter in filters.items()}
关于python - 检查列表中的多个值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54299147/