Python 将列表中的重复项转换为唯一项

我在列表中有以下重复项目列表：

x = ['John', "Jill", "Ron", "John", "Jill", "John", "Tom", "Harry", "Harry"]

我希望获得以下输出：

out = ['John_1', "Jill_1", "Ron_1", "John_2", "Jill_2", "John_3", "Tom_1", "Harry_1", "Harry_2"]

我编写了以下代码：

from collections import Counter
def iterate_string_duplicates(x):
    nl = []
    dct = dict(Counter(x))
    for k,v in dct.items():
        for i in list(range(1,v+1)):
            nl.append(k+"_"+str(i))
    return nl

但由于使用字典方法，我得到了以下输出：

iterate_string_duplicates(x)
output: ['John_1', 'John_2', 'John_3', 'Jill_1', 'Jill_2', 'Ron_1', 'Tom_1', 'Harry_1', 'Harry_2']

---

我正在寻找一种初始字符串顺序不会改变的方法。

您可以使用的列表理解思想是将计数器与每个名称关联起来。那么它只需要一次通过iterable

>>> from collections import defaultdict
>>> from itertools import count
>>> suffix = defaultdict(lambda: count(1))
>>> x = ['John', "Jill", "Ron", "John", "Jill", "John", "Tom", "Harry", "Harry"]
>>> [f"{name}_{next(suffix[name])}" for name in x]
['John_1',
 'Jill_1',
 'Ron_1',
 'John_2',
 'Jill_2',
 'John_3',
 'Tom_1',
 'Harry_1',
 'Harry_2']

与答案类似，我们可以使用从计数1开始的值来确定名称的下一个计数

from collections import defaultdict

names = ["John", "Jill", "Ron", "John", "Jill", "John", "Tom", "Harry", "Harry"]

counts = defaultdict(lambda : 1)

result = []
for name in names:
    result.append(f"{name}_{counts[name]}")
    counts[name] += 1

print(result)

或在添加到结果列表之前保留默认值0和增量：

counts = defaultdict(int)

result = []
for name in names:
    counts[name] += 1
    result.append(f"{name}_{counts[name]}")

与第一个解决方案不同，它将在计数中反映名称的实际计数

输出：

---

您可以使用普通dict执行以下代码并完成任务-

x = ['John', "Jill", "Ron", "John", "Jill", "John", "Tom", "Harry", "Harry"]
y = dict()
def get_me_number(item):
    if item in y:
        y[item] += 1
        return y[item]
    elif item not in y:
        y[item] = 1
        return 1
res = list(map(lambda x_item: x_item+str(get_me_number(x_item)),x))
print(res)

x = ['John', "Jill", "Ron", "John", "Jill", "John", "Tom", "Harry", "Harry"]
y = dict()
def get_me_number(item):
    if item in y:
        y[item] += 1
        return y[item]
    elif item not in y:
        y[item] = 1
        return 1
res = list(map(lambda x_item: x_item+str(get_me_number(x_item)),x))
print(res)