如何计算列表中元素的频率？

Question

我需要在列表中找到元素的频率

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]

输出 - >

b = [4,4,2,1,2]

另外我想从a中删除重复项

a = [1,2,3,4,5]

Answer 1

在Python 2.7(或更新版本)中,您可以使用collections.Counter:

import collections
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counter=collections.Counter(a)
print(counter)
# Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1})
print(counter.values())
# [4, 4, 2, 1, 2]
print(counter.keys())
# [1, 2, 3, 4, 5]
print(counter.most_common(3))
# [(1, 4), (2, 4), (3, 2)]

如果您使用的是Python 2.6或更早版本,可以在此处下载.

Answer 2

由于列表是订购的,您可以这样做:

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
from itertools import groupby
[len(list(group)) for key, group in groupby(a)]

输出:

[4, 4, 2, 1, 2]

Answer 3

Python 2.7+引入了词典理解.从列表中构建字典将获得计数以及删除重复项.

>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[4, 4, 2, 1, 2]

Answer 4

要计算出现次数:

from collections import defaultdict

appearances = defaultdict(int)

for curr in a:
    appearances[curr] += 1

要删除重复项:

a = set(a) 

Answer 5

计算元素的频率可能最好用字典来完成:

b = {}
for item in a:
    b[item] = b.get(item, 0) + 1

要删除重复项,请使用以下集:

a = list(set(a))

Answer 6

在Python 2.7+中,您可以使用collections.Counter来计算项目

>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]

Answer 7

这是另一个简洁的替代方法itertools.groupby，它也适用于无序输入：

from itertools import groupby

items = [5, 1, 1, 2, 2, 1, 1, 2, 2, 3, 4, 3, 5]

results = {value: len(list(freq)) for value, freq in groupby(sorted(items))}

结果

{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}

Answer 8

你可以这样做:

import numpy as np
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
np.unique(a, return_counts=True)

输出:

(array([1, 2, 3, 4, 5]), array([4, 4, 2, 1, 2], dtype=int64))

第一个数组是值,第二个数组是具有这些值的元素数.

所以如果你想得到数字的数组,你应该使用这个:

np.unique(a, return_counts=True)[1]

Answer 9

seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.

Answer 10

from collections import Counter
a=["E","D","C","G","B","A","B","F","D","D","C","A","G","A","C","B","F","C","B"]

counter=Counter(a)

kk=[list(counter.keys()),list(counter.values())]

pd.DataFrame(np.array(kk).T, columns=['Letter','Count'])

Answer 11

我将以以下方式简单地使用scipy.stats.itemfreq：

from scipy.stats import itemfreq

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]

freq = itemfreq(a)

a = freq[:,0]
b = freq[:,1]

您可以在此处查看文档：http : //docs.scipy.org/doc/scipy-0.16.0/reference/generation/scipy.stats.itemfreq.html

Answer 12

数据。假设我们有一个列表：

fruits = ['banana', 'banana', 'apple', 'banana']

解决方案。然后我们可以通过执行以下操作找出列表中每种水果的数量：

import numpy as np    
(unique, counts) = np.unique(fruits, return_counts=True)
{x:y for x,y in zip(unique, counts)}

输出：

{'banana': 3, 'apple': 1}