如何在列表中找到相同的值并将新列表组合在一起? [英] How can I find same values in a list and group together a new list?

问题描述

从此列表中:

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]

我正在尝试创建:

L = [[1],[2,2],[3,3,3],[4,4,4,4],[5,5,5,5,5]]

任何发现相同的值都被分组到它自己的子列表中.到目前为止，这是我的尝试，我想我应该使用 while 循环吗?

Any value which is found to be the same is grouped into it's own sublist. Here is my attempt so far, I'm thinking I should use a while loop?

global n

n = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5] #Sorted list
l = [] #Empty list to append values to

def compare(val):
   """ This function receives index values
   from the n list (n[0] etc) """
   
   global valin
   valin = val

   global count
   count = 0

    for i in xrange(len(n)):
        if valin == n[count]: # If the input value i.e. n[x] == n[iteration]
            temp = valin, n[count]
             l.append(temp) #append the values to a new list
             count +=1
        else:
          count +=1
    

for x in xrange (len(n)):
    compare(n[x]) #pass the n[x] to compare function

推荐答案

有人提到了 N=[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5,5, 5, 5, 1] 会得到 [[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5,5, 5, 5, 5], [1]]

Someone mentions for N=[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 1] it will get [[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5], [1]]

换句话说，当列表的数字不按顺序或列表混乱时，它不可用.

In other words, when numbers of the list isn't in order or it is a mess list, it's not available.

所以我有更好的答案来解决这个问题.

So I have better answer to solve this problem.

from collections import Counter

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]
C = Counter(N)

print [ [k,]*v for k,v in C.items()]

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