Question

我想通过告知2个元素重复多少次来缩短具有重复元素的列表。

list1 = ["New York", "California", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Indiana"]

repetitives = []

for num, element in enumerate(list1):
    if element == list1[num - 2]:
        repetitives.append(element)

core_repetitives = repetitives[0:2]

string_repetitives = ",".join(repetitives)
string_core_repetitives = ",".join(core_repetitives)

repetitives_times = string_repetitives.count(string_core_repetitives)

string_list1 = ",".join(list1)

print string_list1.replace(string_repetitives, "(" + "-".join(core_repetitives) + ") " + str(repetitives_times) + " times")

输出结果为：

New York,California,(Illinois-Texas) 4 times,Illinois,Texas,Indiana

显然它错过了1次。

问题是列表＆＃34;重复＆＃34;没有得到正确的部分，从第34行开始;如果element == list1 [num - 2]：＆＃34;。

通过如何正确认识＆＃34;伊利诺伊州 - 德克萨斯州＆＃34;重复了5次？

谢谢。

相关问题

上述问题是已知的2个要素。但是如果重复部分是重复数量不明的组合怎么办？

例如：

list2 = ["New York", "California", "Illinois", "Texas", "Indiana", "Ohio", "North Carolina", "Washington", "Illinois", "Texas", "Indiana", "Ohio", "North Carolina", "Washington", "Colorado", "Michigan"]

如何分辨[＆＃34; Illinois＆＃34;，＆＃34; Texas＆＃34;，＆＃34; Indiana＆＃34;，＆＃34; Ohio＆＃34;，＆＃34; North Carolina＆＃ 34;，＆＃34;华盛顿＆＃34;]在这里重复了两次？

再次感谢。

Answer 1

以下是我将如何实现您的代码：

from collections import OrderedDict


def repeats(lst):
        return [el for el in lst if lst.count(el) > 1]

def shorten(lst):
    repeat_els = repeats(lst)
    new_lst = [el for el in lst if el not in repeat_els]

    repeats_str = '-'.join(repeat_els)
    core_repeats = '-'.join(list(OrderedDict.fromkeys(repeat_els)))

    repeat_times = repeats_str.count(core_repeats)
    first_repeat_index = lst.index(repeat_els[0])
    repeats_str = '({}) {}'.format(core_repeats, repeat_times)

    new_lst.insert(first_repeat_index, repeats_str)
    return ','.join(new_lst)

概括地说：上面的代码首先将重复和非重复的元素分成两个单独的列表。然后，它将重复的元素格式化为正确的字符串格式，将格式化的字符串添加到非重复元素列表中的正确位置，然后将整个非重复元素列表','.join编辑在一起。

这是一个演示：

>>> list1 = ["New York", "California", "Illinois", 
...          "Texas", "Illinois", "Texas", "Illinois", 
...          "Texas", "Illinois", "Texas", "Illinois", 
...         "Texas", "Indiana"]
>>> 
>>> shorten(list1)
'New York,California,(Illinois-Texas) 5,Indiana'
>>> 
>>> list2 = ["New York", "California", "Illinois", 
...         "Texas", "Indiana", "Ohio", 
...         "North Carolina", "Washington", "Illinois", 
...         "Texas", "Indiana", "Ohio", 
...         "North Carolina", "Washington", "Colorado", 
...         "Michigan"]
>>> shorten(list2)
'New York,California,(Illinois-Texas-Indiana-Ohio-North Carolina-Washington) 2,Colorado,Michigan'
>>>

Answer 2

我想到了一种操纵第一次尝试的方法，让它看起来更好......

笨拙并且不是真正的技术。

即使它看起来不错，但实际上它是错的 - 无论它出现在哪里（伊利诺伊州 - 德克萨斯州）都是额外的（但是它应该只考虑（伊利诺伊州 - 德克萨斯州）从第一次尝试时错过的时间。） / p>

list1 = ["New York", "California", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Illinois", "Texas", "Indiana"]

repetitives = []

for num, element in enumerate(list1):
    if element == list1[num - 2]:
        repetitives.append(element)

core_repetitives = repetitives[0:2]

string_repetitives = ",".join(repetitives)
string_core_repetitives = ",".join(core_repetitives)

repetitives_times = string_repetitives.count(string_core_repetitives)

string_list1 = ",".join(list1)

first_try = string_list1.replace(string_repetitives, "(" + "-".join(core_repetitives) + ") " + str(repetitives_times) + " times")

extra_count = first_try.count(string_core_repetitives)

actual_times = repetitives_times + extra_count

second_try = string_list1.replace(string_repetitives, "(" + "-".join(core_repetitives) + ") " + str(actual_times) + " times")

print second_try.replace(string_core_repetitives, "").replace(",,", ",")

输出是：

New York,California,(Illinois-Texas) 5 times,Indiana

Answer 3

这会将您的字词与其在列表中出现的内容进行映射

from collections import Counter
occurrences = Counter(list1)

然后您可以基于它创建新地图

sublists = {}
for k, v in occurrences.iteritems():
    sublists.setdefault(v, []).append(k)

Python，list有重复的部分

3 个答案: