python字典理解迭代器 [英] python dictionary comprehension iterator

问题描述

 创建一个元素列表
 ＃使用一个字典来查找每个元素的频率
 list = [1,2,6,3,4,5,1,1,3,2,2,5] 
 list.sort （）
 dict = {i：list.count（i）for i in list} 
 print（dict）
  

在字典压缩方法中，for i in list是提供给方法的顺序吗？所以需要1,2,3,4 ..作为关键。我的问题是为什么不需要三次？因为我已经说我在列表中，它不需要把列表中的每一个元素作为一个键？

（我是新的对于python这么容易我！）

解决方案

是的，你的怀疑是正确的。 1 将在迭代中出现3次。然而，由于词典具有唯一的键，所以每次 1 出现时，它将替换以前生成的键/值对与新生成的键/值对。这将给出正确的答案，这不是最有效率的。您可以将列表转换为一个集合，以避免重新处理重复的密钥：

  dict = {i：list.count（i）对于我在set（列表）} 
  

但是，即使这种方法是非常低效的，因为它完全通过列表中每个值的列表，即O（n²）总比较。你可以在列表中单次通过，但是你不会使用dict的理解：

  xs = [1 ，2,6,3,4,5,1,1,3,2,2,5] 
计数= {} 
 x中的x：
计数[x] =计数.get（x，0）+ 1 
  

的结果计数是： {1：3，2：3，3：2，4：1，5：2，6：1}

编辑：我没有意识到图书馆里有什么可以为你做这件事。您应该使用Rohit Jain的解决方案，使用 collections.Counter 。

Hey I have a doubt in the following python code i wrote :

#create a list of elements
#use a dictionary to find out the frequency of each element
list = [1,2,6,3,4,5,1,1,3,2,2,5]
list.sort()
dict = {i: list.count(i) for i in list}
print(dict)

In the dictionary compression method, "for i in list" is the sequence supplied to the method right ? So it takes 1,2,3,4.. as the keys. My question is why doesn't it take 1 three times ? Because i've said "for i in list", doesn't it have to take each and every element in the list as a key ?

(I'm new to python so be easy on me !)

解决方案

Yes, your suspicion is correct. 1 will come up 3 times during iteration. However, since dictionaries have unique keys, each time 1 comes up it will replace the previously generated key/value pair with the newly generated key/value pair. This will give the right answer, it's not the most efficient. You could convert the list to a set instead to avoid reprocessing duplicate keys:

dict = {i: list.count(i) for i in set(list)}

However, even this method is horribly inefficient because it does a full pass over the list for each value in the list, i.e. O(n²) total comparisons. You could do this in a single pass over the list, but you wouldn't use a dict comprehension:

xs = [1,2,6,3,4,5,1,1,3,2,2,5]
counts = {}
for x in xs:
  counts[x] = counts.get(x, 0) + 1

The result for counts is: {1: 3, 2: 3, 3: 2, 4: 1, 5: 2, 6: 1}

Edit: I didn't realize there was something in the library to do this for you. You should use Rohit Jain's solution with collections.Counter instead.

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