使用 python 的平面列表层次结构字符串
Flat list hierachy string using python
如何平仓:
paths = ['[a, c, e]', '[[a, c], [a, b], d]', 'z']
收件人:
paths = [[a, c, e, z], [a, c, d, z] [a, b, d, z]]
- 首先,您需要将每个项目转换为列表示例的表示形式:
'["a", "c", "e"]'
。为此:您应该使用 regex
:
import re
s = '[[a, c], [a, b], d]'
s = re.sub(r'(?<![\]\[]),', '",', re.sub(r',(?![\]\[])',',"',re.sub(r'\s*,\s*', ',', re.sub(r'(?<!\])\]', '"]',re.sub(r'\[(?!\[)', '["', s))))
- 最后,压扁的过程可以用递归算法来描述:
def flat(x):
if not any(isinstance(t, list) == True for t in x):
return x
xx = []
for item in x:
if isinstance(item, str):
for j in range(len(xx)):
xx[j].append(item)
else:
item_new = flat(item)
if any(isinstance(t, list) == True for t in item_new):
xx.extend(item_new)
else:
xx.append(item_new)
return xx
如何平仓:
paths = ['[a, c, e]', '[[a, c], [a, b], d]', 'z']
收件人:
paths = [[a, c, e, z], [a, c, d, z] [a, b, d, z]]
- 首先,您需要将每个项目转换为列表示例的表示形式:
'["a", "c", "e"]'
。为此:您应该使用regex
:
import re
s = '[[a, c], [a, b], d]'
s = re.sub(r'(?<![\]\[]),', '",', re.sub(r',(?![\]\[])',',"',re.sub(r'\s*,\s*', ',', re.sub(r'(?<!\])\]', '"]',re.sub(r'\[(?!\[)', '["', s))))
- 最后,压扁的过程可以用递归算法来描述:
def flat(x):
if not any(isinstance(t, list) == True for t in x):
return x
xx = []
for item in x:
if isinstance(item, str):
for j in range(len(xx)):
xx[j].append(item)
else:
item_new = flat(item)
if any(isinstance(t, list) == True for t in item_new):
xx.extend(item_new)
else:
xx.append(item_new)
return xx