关于python：从dict和list创建sys.path的目录树

Make directory tree of sys.path from dict and list

我试图理解sys.path。

所以我想生成返回目录树的代码，但是我不能。

有人能告诉我密码吗？

路径系统

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['C:\\Users\\81802\\PycharmProjects\\PlayGround',

'C:\\Users\\81802\\AppData\\Local\\Programs\\Python\\Python37\\python37.zip',

'C:\\Users\\81802\\AppData\\Local\\Programs\\Python\\Python37\\DLLs',

'C:\\Users\\81802\\AppData\\Local\\Programs\\Python\\Python37\\lib',

'C:\\Users\\81802\\AppData\\Local\\Programs\\Python\\Python37',

'C:\\Users\\81802\\PycharmProjects\\PlayGround\\venv',

'C:\\Users\\81802\\PycharmProjects\\PlayGround\\venv\\lib\\site-packages',

'C:\\Users\\81802\\PycharmProjects\\PlayGround\\venv\\lib\\site-packages\\setuptools-39.1.0-py3.7.egg',

'C:\\Users\\81802\\PycharmProjects\\PlayGround\\venv\\lib\\site-packages\\pip-10.0.1-py3.7.egg']

【目录树(dict)】

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{'C:\\Users\\81802\':
[{'PycharmProjects\\PlayGround\':
['',
{'venv\':
['',
{'lib\\site-packages\':
['',
'setuptools-39.1.0-py3.7.egg',
'pip-10.0.1-py3.7.egg']}]}]},
{'AppData\\Local\\Programs\\Python\\Python37\':
['',
'python37.zip',
'DLLs',
'lib']}]}

相关讨论

这是我能得到的最简单的。其想法是保持一组目前没有分歧的路径。

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import sys
from pprint import pprint
pprint(sys.path)

sep = '\'

# check if all paths agree on the current name
def isSameName(paths, index):
for path in paths:
if index >= len(path) or path[index] != paths[0][index]:
return False
return True

#transform the current set of paths into tree
def toTree(paths, startIndex):
index = startIndex
if len(paths) == 1:
return sep.join(paths[0][index:])
while isSameName(paths, index):
index += 1
nameMap = dict()
for path in paths:
name = path[index] if len(path) > index else 0
if not (name in nameMap):
nameMap[name] = []
nameMap[name].append(path)
res = [toTree(paths, index) for paths in nameMap.values()]
return { sep.join(paths[0][startIndex:index]) : res}

paths = [path.split(sep) for path in sys.path]
pprint(toTree(paths, 0))

这将为您提供一个字典，其中每个键都是一个目录，值是文件名或带有子目录的字典的列表。

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import os

def get_files_dict(startpath):
tree = [] # this is the array of subdirectory objects
for item in os.listdir(startpath):
# we need to have a full path to the item in the directory
item_path = os.path.join(startpath, item)
if os.path.isfile(item_path):
tree.append(item)
else:
# we call this function recursively for subdirectories
tree.append(get_files_dict(item_path))
return {os.path.basename(startpath):tree}

file_tree = get_files_dict(os.getcwd())

# this is just a helper function to print the tree nicely
def print_tree(d,i=0):
for k,v in d.items():
print("{}{}".format(""*4*i, k+os.sep))
for l in v:
if type(l) is dict:
print_tree(l,i+1)
else:
print("{}{}".format(""*4*(i+1), l))

print_tree(file_tree)

打印输出：

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runner/
.bashrc
.bash_logout
.profile
.site-packages/
main.py
.config/
pycodestyle
_test_runner.py

这是受到这个问题的启发，但我对实现做了很大的改变。