列出重复项的操作答案

【问题标题】：List Manipulation of Duplicates列出重复项的操作
【发布时间】：2019-02-23 05:39:30
【问题描述】：

所以我在一个列表中有我想要操作的元素，基本上我希望这发生：

input:
my_list = ['Gold Trophy (January)', 'Gold Trophy (February)', 'Bronze Trophy 
(March)']

output:
['Gold Trophy x2', 'Bronze Trophy (March)']

当有重复的公共字符串时（例如在 Gold Trophy 的情况下），我希望删除这两个元素，并形成一个新元素，上面写着 (Gold Trophy x(amount of duplicates))

这是我目前所拥有的：

my_list = ['Gold Trophy (January)', 'Gold Trophy (February)', 'Bronze Trophy 
(March)']

# function to count how many duplicates
def countX(my_list, myString): 
    count = 0
    for ele in my_list: 
        if (myString in ele): 
            count = count + 1
    return count 

myString = 'Gold Trophy'
real_count = (countX(my_list, myString))


print(*my_list, sep=', ')
print('duplicates = '+str(countX(my_list, myString)))

此时，此代码运行并返回指定字符串的重复项在列表中的数量。关于从哪里获得所需输出的任何想法？谢谢！

【问题讨论】：

标签： python string list function count

【解决方案1】：

这应该可以在不使用正则表达式的情况下工作。为了清楚地说明正在发生的事情，我已经把 cmets 放在了上面。

from collections import Counter
my_list = ['Gold Trophy (January)', 'Gold Trophy (February)', 'Bronze Trophy (March)']
output_ls = []
trophy_ls = []
month_ls = []
trophy_cnt_dc = {}
for item in my_list:
    trophy_ls.append(item.split(' (')[0])
    month_ls.append(item.split(' (')[1])
# print(trophy_ls) >> ['Gold Trophy', 'Gold Trophy', 'Bronze Trophy']
# print(month_ls) >> ['January)', 'February)', 'March)']
trophy_cnt_dc = dict(Counter(trophy_ls))
#print(trophy_cnt_dc) >> {'Gold Trophy': 2, 'Bronze Trophy': 1}
for k,v in trophy_cnt_dc.items():
    if v > 1:
        output_ls.append(k+' x'+str(v))
    else:
        ind = trophy_ls.index(k)
        output_ls.append(k+' ('+month_ls[ind])
print(output_ls)

输出：

['Gold Trophy x2', 'Bronze Trophy (March)']

【讨论】：

【解决方案2】：

这是一个解决方案（请参阅 cmets 进行澄清）。请注意，我使用了一个小技巧来拆分名称和日期：我在 ( 上拆分，然后在需要时恢复它。可以做得更干净，但不清楚是否需要。

my_list = ['Gold Trophy (January)', 'Gold Trophy (February)', 'Bronze Trophy (March)']

# Create map of tuples: (name, date)
pairs = [tuple(x.split('(')) for x in my_list]

# count the number of each name
counts = dict()
for (name, day) in pairs:
    counts[name] = counts.get(name, 0) + 1

# create a dictionary from initial list
# it doesn't matter how collisions are resolved
# the dictionary is required to process each name only once
init = dict(pairs)
res = []

# for each name:
#   if count is > 1, print the count
#   if count is 1, then print its date
for (name, date) in init.items():
    if counts[name] > 1:
        res.append(name + 'x' + str(counts[name]))
    else:
        res.append(name + '(' + date)
print(res)

【讨论】：