Python os模块找不到文件

Question

与python文件mr robot和subs在同一个目录中有两个文件夹。我有四个文件，在两个文件中都有四个不同的名字。我将文件名存储在mr robot中，列表中没有扩展名(file_s)。我想用列表中的名称重命名subs目录中的文件。但当我运行它时，它会抛出以下错误：

FileNotFoundError: [WinError 2] The system cannot find the file specified

这是我的代码：

import os

currdir = os.getcwd()
vidfiles = os.listdir(f'{currdir}/mr robot')
subfiles = os.listdir(f'{currdir}/subs')
file_s = []


for file in vidfiles:
    filename, _ = os.path.splitext(file)
    file_s.append(filename)


for filename in file_s:

    for sub in subfiles:

        os.rename(sub, f'{filename}.srt')

Answer 1

我认为问题在于，您没有将子文件的全部路径提供给os.rename。

此代码适用于我：

import os

currdir = os.getcwd()
mr_robotpath = f'{currdir}/mr robot'
subspath = f'{currdir}/subs'

vidfiles = os.listdir(mr_robotpath)
subfiles = os.listdir(subspath)

file_s = []

for file in vidfiles:
    filename, _ = os.path.splitext(file)
    file_s.append(filename)

for sub,filename in zip(subfiles,file_s):
    os.rename(f"{subspath}/{sub}", f'{subspath}/{filename}.srt')

Answer 2

我对您的代码做了一些修改，以显示正在发生的事情：

import os

# setup
print("trying to create some files")
try:
    os.mkdir("mr robot")
    os.mkdir("subs")
    for f in "one two three four".split():
        print(f)
        with open("mr robot/"+f+".vid", "w") as h:
            pass
        with open("subs/"+f+".sth", "w") as h:
            pass
except FileExistsError as e:
    print("already there")
    print("(Error: {})".format(e))

currdir = os.getcwd()
vidfiles = os.listdir(f'{currdir}/mr robot')
subfiles = os.listdir(f'{currdir}/subs')
file_s = []

#debug
print("vidfiles", vidfiles)
print("subfiles", subfiles)


for file in vidfiles:
    filename, _ = os.path.splitext(file)
    file_s.append(filename)

print("file_s", file_s)

for filename in file_s:
    print("filename", filename)

    for sub in subfiles:
        newname = f'{filename}.srt'
        print(f"    {sub} -> {newname}")
        # os.rename(sub, newname)

它的产出如下：

trying to create some files
already there
(Error: [WinError 183] Cannot create a file when that file already exists: 'mr robot')
vidfiles ['four.vid', 'one.vid', 'three.vid', 'two.vid']
subfiles ['four.sth', 'one.sth', 'three.sth', 'two.sth']
file_s ['four', 'one', 'three', 'two']
filename four
    four.sth -> four.srt
    one.sth -> four.srt
    three.sth -> four.srt
    two.sth -> four.srt
filename one
    four.sth -> one.srt
    one.sth -> one.srt
    three.sth -> one.srt
    two.sth -> one.srt
filename three
    four.sth -> three.srt
    one.sth -> three.srt
    three.sth -> three.srt
    two.sth -> three.srt
filename two
    four.sth -> two.srt
    one.sth -> two.srt
    three.sth -> two.srt
    two.sth -> two.srt

正如所怀疑的(请参阅注释)，您有两个循环，结果是4*4重命名。

修改后的代码更接近于您试图实现的目标：

import os

currdir = os.getcwd()
vidfiles = os.listdir(f'{currdir}/mr robot')
subfiles = os.listdir(f'{currdir}/subs')
file_s = []

# debug
print("vidfiles", vidfiles)
print("subfiles", subfiles)


for file in vidfiles:
    filename, _ = os.path.splitext(file)
    file_s.append(filename)

print("file_s", file_s)

for i,sub in enumerate(subfiles):
    newname = f'{file_s[i]}.srt'
    print(f"    {sub} -> {newname}")
    # os.rename(sub, newname)

印出来

vidfiles ['four.vid', 'one.vid', 'three.vid', 'two.vid']
subfiles ['four.sth', 'one.sth', 'three.sth', 'two.sth']
file_s ['four', 'one', 'three', 'two']
    four.sth -> four.srt
    one.sth -> one.srt
    three.sth -> three.srt
    two.sth -> two.srt

但是，由于不存在从视频文件到字幕文件的映射，您可能会得到错误的顺序，因为即使按字母顺序排序会导致文件匹配：

https://docs.python.org/3/library/os.html#os.listdir

返回一个列表，其中包含路径给出的目录中条目的名称。列表是按任意顺序排列的，不包括特殊条目'.‘。还有“..”即使它们存在于目录中。

所以你应该自己整理一下：

• https://docs.python.org/3/howto/sorting.html#sortinghowto
• https://docs.python.org/3/library/stdtypes.html#list.sort (in place)
• https://docs.python.org/3/library/functions.html#sorted (新列表)

)

​

此外，另一个答案是正确的，因为您还需要提供文件的路径(或更改工作目录)，例如：

​

os.rename("subs/"+sub, "subs/"+newname)