Question

我有一个清单：

list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']

我想创建一个包含如下项目数的新列表：

list_b = ['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

我尝试过这样的事情：

list_b=[]
for item in list_a:
    list_b.append(item+"_"+str(list_a.count(item)))

但这当然会增加每个元素的总计数。

Answer 1

您可以使用dict存储索引：

list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']

d={}
r=[]
for i in list_a:
    d.setdefault(i, 0)
    d[i]+=1
    r.append(i+"_"+str(d[i]))

print r

输出：

['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

另一种Pythonic方式：

>>> list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']
>>> d={}
>>> [i+"_"+str(len(d[i])) for i in list_a if not d.setdefault(i,[]).append(True)]
['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

Answer 2

您可以使用enumerate，

执行此操作

[j+'_'+str(list_a[:i+1].count(j)) for i,j in enumerate(list_a)]

<强>观

使用enumerate我也会得到元素的索引，所以会切入当前元素并计算出现的数量切片清单。

<强>结果：

['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

执行时间：

根据我在回答关于执行时间的评论中的讨论，我已经完成了这里实现的所有方法，以下是时间，

In [68]: %timeit Mc_grady_method_1()
100000 loops, best of 3: 4.29 µs per loop

In [69]: %timeit Mc_grady_method_2()
100000 loops, best of 3: 4.35 µs per loop

In [70]: %timeit Rahul_KP()
100000 loops, best of 3: 3.8 µs per loop

In [71]: %timeit Moe_A()
100000 loops, best of 3: 3.94 µs per loop

In [72]: %timeit Allen()
100000 loops, best of 3: 13.1 µs per loop

In [73]: %timeit Mayur_Buragohain()
100000 loops, best of 3: 3.86 µs per loop

In [74]: %timeit Martin_Evans()
100000 loops, best of 3: 10.5 µs per loop

我的方法仍然在这方面取得了不错的表现。

Answer 3

使用Python的Counter()计算每个单词：

from collections import Counter

word_count = Counter()
list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']
list_b = []

for word in list_a:
    word_count[word] += 1
    list_b.append('{}_{}'.format(word, word_count[word]))

print list_b

给你：

['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

Answer 4

import collections

list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']

#build a counter for each word and a index list for each word.
counter = {k:list(range(v+1,0,-1)) for k,v in collections.Counter(list_a).items()}

#attach index to each each occurance of a word
list_b = [e+'_'+str(counter[e].pop()) for e in list_a]
print(list_b)

['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

Answer 5

检查一下 -

list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']
list_b=[]
someDict={}
for item in list_a:
    if item in someDict.keys():
        temp_count=someDict[item]+1
        temp_item=list_b.append(item+"_"+str(temp_count))
        someDict[item]=temp_count
    else:
        list_b.append(item+"_1")
        someDict[item]=1
print list_b

Answer 6

您可以使用临时list添加单词，然后将其添加到list_b，其计数如下：

list_a = ['hello', 'goodbye', 'goodbye', 'hello', 'whatever', 'whatever', 'whatever', 'hello']
    list_b = []
    tmp = []
    for word in list_a:
        tmp.append(word)
        list_b.append(word + '_' + str(tmp.count(word)))
    print list_b

输出：

['hello_1', 'goodbye_1', 'goodbye_2', 'hello_2', 'whatever_1', 'whatever_2', 'whatever_3', 'hello_3']

Answer 7

如果您知道列表中将包含哪些元素，则可以创建一个变量来计算它们。你甚至可以用两个循环来完成它：

提供的代码并不是最明智的方法，但它应该运行良好

list_items=[]
counters_items=[]
for item in list_a:
    if item in list_items:
        pass
    else:
        list_items.append(item)

# Now we have stored a list of all type of item

list_b = list_a.copy()
for item in list_items:
    counter = 1
    for it in list_b:
        if item == it:
            it = it + "_" + str(counter)
            counter +=1
    # If you want to make sure the whole list has been numbered
    if counter != list_a.count(item) + 1:
        print "Smth wrong happened"

从列表计数出现并创建添加了计数的新列表

7 个答案: