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我正在尝试将列表newtrack（文件夹中的歌曲名称）与电子表格中的列表filezto和相册进行匹配。 如果文件夹中的歌曲名称匹配列表文件ZTO并具有Nasb的唱片集，则将其复制到文件夹列表（ORDFLDTOSRT），如果不是Nasb唱片集，则将歌曲复制到列表（NTSUFFLDRS）

由于某些原因，在运行时，if语句不会同步它们，以便打印它们是否匹配并将文件复制到文件夹中。我尝试了不同的选择，但无法使其工作。提前谢谢

albm = ["NASB", "NASB", "NASB", "NASB", "NTSB", "NTSB", "NTSB", "NTSB"]

newtracks = [
    "Track1.mp3",
    "Track2.mp3",
    "Track3.mp3",
    "Track4.mp3",
    "Track5.mp3",
    "Track5.mp3",
]

filezto = [
    "Track1.mp3",
    "Track2.mp3",
    "Track3.mp3",
    "Track4.mp3",
    "Track5.mp3",
    "Track5.mp3",
]

ordfldrtosrt = ["1-GEN-GENE", "2 - EXO - Exos", "3 - LEV - Lev", "4 - NUM - Numb"]

ntsubfldrs = ["1-MAT-Matt", "2 - MRK - Mar", "3 - LUK - LI", "4 - JHN - Jo"]

for album in albm:
    for newtracks, filezto, ordfldrtosrt, ntsubfldrs in (
        (ntk, ft, otm, nts)
        for ntk in newtracks
        for ft in filezto
        for otm in ordfldrtosrt
        for nts in ntsubfldrs
    ):
        if album == "Nasb" and newtracks == filezto:

            print("Yes", newtracks, " files to", filezto)

            # shutil.copy('C:\\Users\\mrdrj\\Desktop\\SJ\\NASB\\'+newtracks,  'D:\\Desktop\\pythonProject\\Nasb\\'+ ordfldrtosrt+'\\'+newtracks)
        else:
            print("No", newtracks, " files to", filezto)
            # shutil.copy('C:\\Users\\mrdrj\\Desktop\\SJ\\NASB\\'+newtracks,  'D:\\Desktop\\pythonProject\\Nasb\\'+ ntsubfldrs +'\\'+newtracks)

Answer 1

从您上次的评论中，我猜您有一个来自spreadsheet的（专辑、曲目）列表，您正试图将其与folder中一个专辑的曲目列表相匹配：

folder = [
    "Track1.mp3",
    "Track2.mp3",
    "Track3.mp3",
    "Track4.mp3",
    "Track5.mp3",
    "Track5.mp3",
]
spreadsheet = [
    ( "NASB", "Track1.mp3" ),
    ( "NASB", "Track2.mp3" ),
    ( "NASB", "Track3.mp3" ),
    ( "NASB", "Track4.mp3" ),
    ( "NASB", "Track5.mp3" ),
    ( "NASB", "Track5.mp3" ),
    ( "NTSB", "Track1.mp3" ),
    ( "NTSB", "Track2.mp3" ),
    ( "NTSB", "Track3.mp3" ),
    ( "NTSB", "Track4.mp3" ),
    ( "NTSB", "Track5.mp3" ),
    ( "NTSB", "Track5.mp3" )
]
for album, track in spreadsheet:
    if album == 'NASB' and track in folder:
        print('%s %s' % ('Yes', track))
    else:
        print('%s %s' % ('No', track))

with将为您提供以下输出：

Yes Track1.mp3
Yes Track2.mp3
Yes Track3.mp3
Yes Track4.mp3
Yes Track5.mp3
Yes Track5.mp3
No Track1.mp3
No Track2.mp3
No Track3.mp3
No Track4.mp3
No Track5.mp3
No Track5.mp3