for subdir in ('A', 'B', 'C'):
path = 'X/{}/result.txt'.format(subdir)
with open(path, 'w') as fp:
fp.write("This is the result!\n")
或者,也许:
^{pr2}$
更新:
对于注释中列出的附加要求,我建议您使用glob.glob(),如下所示:
import os.path
import glob
for subdir in glob.glob("X/certain*"):
path = os.path.join(subdir, "result.txt")
with open(path, 'w') as fp:
fp.write("This is the result\n")
import os, fnmatch
def modify_file(filepath):
try:
with open(filepath) as f:
s = f.read()
s = your_method(s) # return modified content
if s:
with open(filepath, "w") as f: #write modified content to same file
print filepath
f.write(s)
f.flush()
f.close()
except:
import traceback
print traceback.format_exc()
def modifyFileInDirectory(directory, filePattern):
for path, dirs, files in os.walk(os.path.abspath(directory), followlinks=True):
for filename in fnmatch.filter(files, filePattern):
modify_file(filename)
modifyFileInDirectory(your_path,file_patter)
如果所有文件都有相同的名称(比如,
result.txt
),那么循环相当简单:或者,也许:
^{pr2}$更新:
对于注释中列出的附加要求,我建议您使用
glob.glob()
,如下所示:在上面的例子中,对
glob.glob()
的调用将返回X的所有子目录的列表,这些子目录以文本“searent”开头。所以,这个循环可能会连续地创建“X/certainA/结果.txt“和”X/certainBigHouse/结果.txt,假设这些子目录已经存在。在要修改遍历其子目录的不同文件的内容,可以使用手术室步行和fnmatch-选择要修改的正确文件格式!在
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