Question

我有两个列表，我想在列表中返回相似值的len（）。

A = [1,1,2,2]
B = [3,3,3,3,7,7,7]

在第一个列表中有两次数字1和2，我想在列表中使用len of number值，以查看数字1在第一个列表中重复多少次。在这种情况下，数字2将是2，数字2将是2。

Answer 1

这是collections.Counter

的工作

>>> from collections import Counter
>>> Counter([1,1,2,2])
Counter({1: 2, 2: 2})
>>> Counter([3,3,3,3,7,7,7])
Counter({3: 4, 7: 3})

Answer 2

一种不使用集合计数器的单行解决方案。

A=[3,4,4,4,3,5,6,8,4,3]
duplicates=dict(set((x,A.count(x)) for x in filter(lambda rec : A.count(rec)>1,A)))
output:
{3: 3, 4: 4}

但是，此解决方案不考虑“拉伸”

Answer 3

您可以简单地遍历数字并计算相同的数字-或使用itertools.groupby：

def count_em(l):
    """Returns a list of lenghts of consecutive equal numbers as list. 
    Example: [1,2,3,4,4,4,3,3] ==> [1,1,1,3,2]"""
    if not isinstance(l,list):
        return None

    def count():
        """Counts equal elements, yields each count"""
        # set the first elem as current
        curr = [l[0]]

        # for the rest of elements
        for elem in l[1:]:
            if elem == curr[-1]:
                # append as long as the element is same as last one in curr 
                curr.append(elem)
            else:
                # yield the number
                yield len(curr)
                # reset curr to count the new ones
                curr = [elem]
        # yield last group
        yield len(curr)

    # get all yields and return them as list
    return list(count())


def using_groupby(l):
    """Uses itertools.groupby and a list comp to get the lenghts."""
    from itertools import groupby
    grp = groupby(l) # this groups by the elems themselfs
    # count the grouped items and return as list
    return [ sum(1 for _ in items) for g,items in grp]

测试：

A = [1,1,2,2]
B = [3,3,3,3,7,7,7]
C = [1,1,2,2,2,1,1,1,1,1,6,6]

for e in [A,B,C]:
    print(count_em(e),  using_groupby(e))

输出：

# count_em     using_groupby    Input
[2, 2]         [2, 2]         # [1,1,2,2]
[4, 3]         [4, 3]         # [3,3,3,3,7,7,7]
[2, 3, 5, 2]   [2, 3, 5, 2]   # [1,1,2,2,2,1,1,1,1,1,6,6]

返回列表中相似值的len

3 个答案: