[英]Python - make non-unique items in list unique by adding count
How would I make the items in my list unique by concatenating a count, starting from 1 for each unique value?如何通过连接计数使列表中的项目变得唯一,每个唯一值从 1 开始?
so, for example:所以,例如:
sheep, sheep, tiger, sheep, hippo, tiger
becomes:变成:
sheep1, sheep2, tiger1, sheep3, hippo1, tiger2
Here's how you could use Counter to do it.下面是如何使用 Counter 来做到这一点。
from collections import Counter
s = ["sheep", "sheep", "tiger", "sheep", "hippo", "tiger"]
u = [ f"{a}{c[a]}" for c in [Counter()] for a in s if [c.update([a])] ]
print(u)
['sheep1', 'sheep2', 'tiger1', 'sheep3', 'hippo1', 'tiger2']
note that, if your strings can have a numeric suffix, this would not be sufficient to cover all cases (eg ['alpha']*11+['alpha1']
would repeat 'alpha11'
)请注意,如果您的字符串可以有数字后缀,则这不足以涵盖所有情况(例如
['alpha']*11+['alpha1']
会重复'alpha11'
)
you can use a simple for
loop:你可以使用一个简单的
for
循环:
l = ['sheep', 'sheep', 'tiger', 'sheep', 'hippo', 'tiger']
count = {}
output = []
for s in l:
if s in count:
count[s] += 1
else:
count[s] = 1
output.append(f'{s}{count[s]}')
output
output:输出:
['sheep1', 'sheep2', 'tiger1', 'sheep3', 'hippo1', 'tiger2']
Using a combination of defaultdict
and count
:使用
defaultdict
和count
的组合:
>>> from collections import defaultdict
>>> from itertools import count
>>> s = ["sheep", "sheep", "tiger", "sheep", "hippo", "tiger"]
>>> d = defaultdict(lambda: count(1))
>>> [f'{x}{next(d[x])}' for x in s]
['sheep1', 'sheep2', 'tiger1', 'sheep3', 'hippo1', 'tiger2']
count
is an object that yields ever-increasing numbers as you iterate over it; count
是一个对象,当您对其进行迭代时,它会产生不断增加的数字; calling next
gives you the next number in the sequence.调用
next
为您提供序列中的下一个数字。
The defaultdict
creates a new count
instance every time you try to access a new key, while saving the newly created instance for the next time you see the same key.每次尝试访问新密钥时,
defaultdict
都会创建一个新的count
实例,同时保存新创建的实例以供下次看到相同的密钥时使用。
I had a very similar need where the output would be:我有一个非常相似的需求,输出将是:
['sheep', 'sheep1', 'tiger', 'sheep2', 'hippo', 'tiger1']
I approached it a little differently looking for an O(n) solution and extended the dictionary class.我以不同的方式寻找 O(n) 解决方案并扩展了字典类。
class IncDict(dict):
def __missing__(self,key):
return -1
def __getitem__(self,key):
val = dict.__getitem__(self,key)
val+=1
dict.__setitem__(self,key,val)
if val==0:
return key
else:
return key+str(val)
l = ['sheep', 'sheep', 'tiger', 'sheep', 'hippo', 'tiger']
uniquify = IncDict()
[uniquify[x] for x in l]
Output:输出:
['sheep', 'sheep1', 'tiger', 'sheep2', 'hippo', 'tiger1']
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