[英]How to divide a list in python?
I have this list example list1=[['p1', 'p2', 'p3', 'p4'], ['p5', 'p6', 'p7']]
and if any variable is the same as other variable them split it in a group. 我有这个列表示例
list1=[['p1', 'p2', 'p3', 'p4'], ['p5', 'p6', 'p7']]
并且如果任何变量与其他变量相同他们把它分成一组。 Lets say p1 and p4 are the same and p5 and p6. 让我们说p1和p4是相同的,p5和p6。 So I want a new list that looks like
list2=[['p1', 'p4'], ['p2', 'p3'], ['p5', 'p6'], 'p7']
. 所以我想要一个新列表,看起来像
list2=[['p1', 'p4'], ['p2', 'p3'], ['p5', 'p6'], 'p7']
。 So how I need to divide them, pls help. 所以我需要如何划分它们,请帮忙。 I am using the newest version of python.
我正在使用最新版本的python。
Ok to be more specific for "the same" in my program, I use p1 and p4 and if they give the same result for a certain character then I merge them in one group. 好的,对于我的程序中的“相同”更具体,我使用p1和p4,如果它们为某个字符给出相同的结果,那么我将它们合并到一个组中。 Example
例
if dictionary.get(p1, character) is dictionary.get(p4, character)
if you have more questions just ask me. 如果你有更多问题,请问我。
The following will give you that result: 以下将给出结果:
list1=[[1, 2, 2, 1], [3, 4, 3]]
print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]
[OUTPUT]
[[1, 1], [2, 2], [3, 3], [4]]
list1=[['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]
print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]
[OUTPUT]
[['what'], ['hello', 'hello', 'hello'], ['i', 'i'], ['am']]
list1=[[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]
print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]
[OUTPUT]
[[1, 1], [2, 2], [3], [8, 8], [9, 9], [7], [4, 4], [5, 5], [6]]
Using the unique_everseen
recipe and collections.Counters
: 使用
unique_everseen
配方和collections.Counters
:
from collections import Counter
def solve(lst):
counters = map(Counter, lst)
return [ [uniq]*c[uniq] for seq, c in zip(lst, counters)
for uniq in unique_everseen(seq)]
Demo: 演示:
>>> print(solve([[1, 2, 2, 1], [3, 4, 3]]))
[[1, 1], [2, 2], [3, 3], [4]]
>>> print(solve([['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]))
[['hello', 'hello', 'hello'], ['what'], ['i', 'i'], ['am']]
>>> print(solve([[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]))
[[1, 1], [2, 2], [3], [9, 9], [8, 8], [7], [5, 5], [4, 4], [6]]
As you can see this also preserves the order of the items. 如您所见,这也保留了项目的顺序。
Code for unique_everseen
recipe: unique_everseen
配方的代码:
from itertools import filterfalse
def unique_everseen(iterable, key=None):
"List unique elements, preserving order. Remember all elements ever seen."
# unique_everseen('AAAABBBCCDAABBB') --> A B C D
# unique_everseen('ABBCcAD', str.lower) --> A B C D
seen = set()
seen_add = seen.add
if key is None:
for element in filterfalse(seen.__contains__, iterable):
seen_add(element)
yield element
else:
for element in iterable:
k = key(element)
if k not in seen:
seen_add(k)
yield element
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