如何对以"."分隔的字符串列表进行排序，中间还有数字？

Question

我有一个字符串列表，其中包含由点.分隔的命令，如下所示：

DeviceA.CommandA.1.Hello,
DeviceA.CommandA.2.Hello,
DeviceA.CommandA.11.Hello,
DeviceA.CommandA.3.Hello,
DeviceA.CommandB.1.Hello,
DeviceA.CommandB.1.Bye,
DeviceB.CommandB.What,
DeviceA.SubdeviceA.CommandB.1.Hello,
DeviceA.SubdeviceA.CommandB.2.Hello,
DeviceA.SubdeviceB.CommandA.1.What

我想按自然顺序订购：

1. 顺序必须按字段索引排列优先级(例如，以DeviceA开头的命令将始终位于DeviceB之前等(
2. 按字母顺序排列字符串
3. 当它找到一个按升序数字排序的数字时

因此，排序输出应为：

DeviceA.CommandA.1.Hello,
DeviceA.CommandA.2.Hello,
DeviceA.CommandA.3.Hello,
DeviceA.CommandA.11.Hello,
DeviceA.CommandB.1.Bye,
DeviceA.CommandB.1.Hello,
DeviceA.SubdeviceA.CommandB.1.Hello,
DeviceA.SubdeviceA.CommandB.2.Hello,
DeviceA.SubdeviceB.CommandA.What,
DeviceB.CommandB.What

还请注意，命令字段的长度是动态的，用点分隔的字段数量可以是任何大小。

到目前为止，我没有运气地尝试过(数字按字母顺序排列，例如11在5之前(：

list = [
"DeviceA.CommandA.1.Hello",
"DeviceA.CommandA.2.Hello",
"DeviceA.CommandA.11.Hello",
"DeviceA.CommandA.3.Hello",
"DeviceA.CommandB.1.Hello",
"DeviceA.CommandB.1.Bye",
"DeviceB.CommandB.What",
"DeviceA.SubdeviceA.CommandB.1.Hello",
"DeviceA.SubdeviceA.CommandB.2.Hello",
"DeviceA.SubdeviceB.CommandA.1.What"
]
sorted_list = sorted(list, key=lambda x: x.split('.')) 

编辑：更正了打字错误。

Answer 1

这样的事情应该会让你开始。

from pprint import pprint
data_list = [
"DeviceA.CommandA.1.Hello",
"DeviceA.CommandA.2.Hello",
"DeviceA.CommandA.3.Hello",
"DeviceA.CommandB.1.Hello",
"DeviceA.CommandB.1.Bye",
"DeviceB.CommandB.What",
"DeviceA.SubdeviceA.CommandB.1.Hello",
"DeviceA.SubdeviceA.CommandB.15.Hello",  # added test case to ensure numbers are sorted numerically
"DeviceA.SubdeviceA.CommandB.2.Hello",
"DeviceA.SubdeviceB.CommandA.1.What",
]

def get_sort_key(s):
# Turning the pieces to integers would fail some comparisons (1 vs "What")
# so instead pad them on the left to a suitably long string
return [
bit.rjust(30, "0") if bit.isdigit() else bit
for bit in s.split(".")
]

# Note the key function must be passed as a kwarg.
sorted_list = sorted(data_list, key=get_sort_key)
pprint(sorted_list)

输出为

['DeviceA.CommandA.1.Hello',
'DeviceA.CommandA.2.Hello',
'DeviceA.CommandA.3.Hello',
'DeviceA.CommandB.1.Bye',
'DeviceA.CommandB.1.Hello',
'DeviceA.SubdeviceA.CommandB.1.Hello',
'DeviceA.SubdeviceA.CommandB.2.Hello',
'DeviceA.SubdeviceA.CommandB.15.Hello',
'DeviceA.SubdeviceB.CommandA.1.What',
'DeviceB.CommandB.What']

Answer 2

在sorted中指定key似乎可以实现您想要的：

import re
def my_key(s):
n = re.search("d+",s)
return (s[:n.span()[0]], int(n[0])) if n else (s,) 
print(sorted(l, key = my_key))

输出：

['DeviceA.CommandA.1.Hello', 'DeviceA.CommandA.2.Hello', 'DeviceA.CommandA.3.Hello', 'DeviceA.CommandA.11.Hello', 'DeviceA.CommandB.1.Hello', 'DeviceA.CommandB.1.Bye', 'DeviceA.SubdeviceA.CommandB.1.Hello', 'DeviceA.SubdeviceA.CommandB.2.Hello', 'DeviceA.SubdeviceB.CommandA.1.What', 'DeviceB.CommandB.What']

Answer 3

有很多方法可以实现这一点。这里有一个不依赖于任何额外模块的进口：

LOS = ['DeviceA.CommandA.1.Hello',
'DeviceA.CommandA.2.Hello',
'DeviceA.CommandA.11.Hello',
'DeviceA.CommandA.3.Hello',
'DeviceA.CommandB.1.Hello',
'DeviceA.CommandB.1.Bye',
'DeviceB.CommandB.What',
'DeviceA.SubdeviceA.CommandB.1.Hello',
'DeviceA.SubdeviceA.CommandB.2.Hello',
'DeviceA.SubdeviceB.CommandA.1.What']

def func(s):
tokens = s.split('.')
for i, token in enumerate(tokens):
try:
v = int(token)
return ('.'.join(tokens[0:i]), v)
except ValueError:
pass
return (s, 0)

print(sorted(LOS, key=func))