如何汇总列表中的连续重复项?

时间:2019-06-11 06:29:58

标签: python python-3.x list

我在互联网上遇到一些实践问题,申请实习并结业。问题是相同的连续数字求和。

示例:[1,1,3,4,4,5] ---> [2,3,8,5]

 def sum_consecutive(s):
    p = []
    for i in range(len(s)):
        if s[i] == s[i-1]:
            p.append(s[i] + s[i-1])
            p.remove(s[i])
        elif s[i] != s[i-1]:
            p.append(s[i])
    return p

在[1,4,4,4,0,4,3,3,1]上运行上述代码时,它应该返回[1,12,0,4,6,1]而不是

sum_conecutive(s)

    if s[i] == s[i-1]:
             p.append(s[i] + s[i-1])
             p.remove(s[i]) #This line is the problem
    elif s[i] != s[i-1]:
             p.append(s[i])

错误:

ValueError: list.remove(x): x not in list

3 个答案:

答案 0 :(得分:5)

使用变量来跟踪最后看到的值,然后将其添加到当前索引(如果相同)。

def sum_consecutive(s):
    p = s[:1]   # final list
    cur = s[0]  # keep track of last seen value

    for i in s[1:]:  # your exercise: replace this with `range` 
        if i == cur:
            p[-1] += i
        else:
            p.append(i)
            cur = i

    return p

sum_consecutive([1, 4, 4, 4, 0, 4, 3, 3, 1])
# [1, 12, 0, 4, 6, 1]

作为奖励,请不要忘记stdlib存在,因此您可以使用itertools.groupby在一行中完成此操作。

from itertools import groupby
[sum(g) for _, g in groupby([1, 4, 4, 4, 0, 4, 3, 3, 1])]
# [1, 12, 0, 4, 6, 1]

答案 1 :(得分:2)

def sum_consecutive(input_list):
i = 0
res = []
while(i < len(input_list)):
    j = i
    while(j < len(input_list) and input_list[i]==input_list[j]):
        j = j + 1

    res.append(input_list[i]*(j-i))
    i = j
return res

答案 2 :(得分:0)

我写了这个功能,它很有趣,因为它很容易理解:

def count_occ(L):
    p = []
    count = 1
    for i, j in zip(L[:-1], L[1:]):
        if i != j:
            p.append(count*i)
            count = 1
        else:
            count += 1
    p.append(count*L[-1])
    return p