如何计算的频率要素的列表?
题
我需要找到频率的因素中的一个列表
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
输出>
b = [4,4,2,1,2]
我也想要删除的重复,从
a = [1,2,3,4,5]
解决方案
由于列表进行排序,你可以这样做:
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]
其他提示
在Python 2.7(或更高版本),则可以使用 collections.Counter
一个>:
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)]
如果您正在使用Python 2.6或以上,你可以下载 rel="noreferrer">这里。
的Python 2.7+介绍字典理解。建立从列表中词典将让你以及摆脱重复的次数。
>>> 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]
要计数出现数:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1
要删除重复:
a = set(a)
计数元件的频率可能是最好用字典完成:
b = {}
for item in a:
b[item] = b.get(item, 0) + 1
要删除重复的,使用一组:
a = list(set(a))
在Python 2.7+,您可以使用 collections.Counter 来计数项
>>> 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]
下面是一个使用itertools.groupby
另一个succint替代这也适用于无序输入:
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))}
<强>结果
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
您可以这样做:
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))
在第一阵列是值,并且所述第二阵列是具有这些值的元素的数量。
所以,如果你想获得你应该使用这个数字只是数组:
np.unique(a, return_counts=True)[1]
seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.
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'])
我只想利用这.统计数据。itemfreq在以下方式:
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]
你可以检查的文件: http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.itemfreq.html
def frequencyDistribution(data):
return {i: data.count(i) for i in data}
print frequencyDistribution([1,2,3,4])
...
{1: 1, 2: 1, 3: 1, 4: 1} # originalNumber: count
有关你的第一个问题,迭代列表并使用字典来保持一个元素的轨道existsence。
有关第二个问题,只需使用集合运算符。
此答案是更明确的
a = [1,1,1,1,2,2,2,2,3,3,3,4,4]
d = {}
for item in a:
if item in d:
d[item] = d.get(item)+1
else:
d[item] = 1
for k,v in d.items():
print(str(k)+':'+str(v))
# output
#1:4
#2:4
#3:3
#4:2
#remove dups
d = set(a)
print(d)
#{1, 2, 3, 4}
我是比较晚,但是这也将工作,并会帮助别人:
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq_list = []
a_l = list(set(a))
for x in a_l:
freq_list.append(a.count(x))
print 'Freq',freq_list
print 'number',a_l
会产生这个..
Freq [4, 4, 2, 1, 2]
number[1, 2, 3, 4, 5]
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
# 1. Get counts and store in another list
output = []
for i in set(a):
output.append(a.count(i))
print(output)
# 2. Remove duplicates using set constructor
a = list(set(a))
print(a)
- 集的集合不允许重复的,通过列表的设置()构造将得到一个迭代的完全独特的对象。count()function返回的整数计时的对象是在一个列表中通过。与的唯一对象的计算和每个数值储存通过附加一个空的名单输出
- list()构造是使用转换设置(a)入名单,并称相同的一个变量
输出
D:\MLrec\venv\Scripts\python.exe D:/MLrec/listgroup.py
[4, 4, 2, 1, 2]
[1, 2, 3, 4, 5]
from collections import OrderedDict
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
def get_count(lists):
dictionary = OrderedDict()
for val in lists:
dictionary.setdefault(val,[]).append(1)
return [sum(val) for val in dictionary.values()]
print(get_count(a))
>>>[4, 4, 2, 1, 2]
要删除重复和维护命令:
list(dict.fromkeys(get_count(a)))
>>>[4, 2, 1]
我使用计数器来生成频率。从文本文件的话字典在1行代码
def _fileIndex(fh):
''' create a dict using Counter of a
flat list of words (re.findall(re.compile(r"[a-zA-Z]+"), lines)) in (lines in file->for lines in fh)
'''
return Counter(
[wrd.lower() for wrdList in
[words for words in
[re.findall(re.compile(r'[a-zA-Z]+'), lines) for lines in fh]]
for wrd in wrdList])
使用字典简单的解决方案。
def frequency(l):
d = {}
for i in l:
if i in d.keys():
d[i] += 1
else:
d[i] = 1
for k, v in d.iteritems():
if v ==max (d.values()):
return k,d.keys()
print(frequency([10,10,10,10,20,20,20,20,40,40,50,50,30]))
#!usr/bin/python
def frq(words):
freq = {}
for w in words:
if w in freq:
freq[w] = freq.get(w)+1
else:
freq[w] =1
return freq
fp = open("poem","r")
list = fp.read()
fp.close()
input = list.split()
print input
d = frq(input)
print "frequency of input\n: "
print d
fp1 = open("output.txt","w+")
for k,v in d.items():
fp1.write(str(k)+':'+str(v)+"\n")
fp1.close()
然而,随着另一个算法的另一个解决方案,而无需使用集合:
def countFreq(A):
n=len(A)
count=[0]*n # Create a new list initialized with '0'
for i in range(n):
count[A[i]]+= 1 # increase occurrence for value A[i]
return [x for x in count if x] # return non-zero count
num=[3,2,3,5,5,3,7,6,4,6,7,2]
print ('\nelements are:\t',num)
count_dict={}
for elements in num:
count_dict[elements]=num.count(elements)
print ('\nfrequency:\t',count_dict)
可以使用在python提供的内置功能
l.count(l[i])
d=[]
for i in range(len(l)):
if l[i] not in d:
d.append(l[i])
print(l.count(l[i])
上面的代码自动在一个列表中删除重复和同时打印在原始列表中的每个元素和无重复列表的频率。
二鸟一重击! X d
此方法可以尝试,如果你不希望使用任何库,并保持它的简单和短!
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
marked = []
b = [(a.count(i), marked.append(i))[0] for i in a if i not in marked]
print(b)
O / P
[4, 4, 2, 1, 2]
为了记录在案,一个功能性的回答:
>>> L = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> import functools
>>> >>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc,1)] if e<=len(acc) else acc+[0 for _ in range(e-len(acc)-1)]+[1], L, [])
[4, 4, 2, 1, 2]
它的清洁,如果你数数零:
>>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc)] if e<len(acc) else acc+[0 for _ in range(e-len(acc))]+[1], L, [])
[0, 4, 4, 2, 1, 2]
一个解释:
- 我们从一个空的
acc
列表; - 如果下一个元素
e
的L
低于尺寸的acc
, 我们只需更新这个元素:v+(i==e)
装置v+1
如果索引i
的acc
是的目前的元素e
, 否则的先前的数值v
; - 如果下一个元素
e
的L
大于或等于要的大小acc
, 我们必须扩大acc
主办新的1
.
元素没有被排(itertools.groupby
).你就变得怪异结果,如果你有负数。
另一种方式是使用一个字典和list.count,下方的幼稚的方式来做到这一点。
dicio = dict()
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
b = list()
c = list()
for i in a:
if i in dicio: continue
else:
dicio[i] = a.count(i)
b.append(a.count(i))
c.append(i)
print (b)
print (c)
a=[1,2,3,4,5,1,2,3]
b=[0,0,0,0,0,0,0]
for i in range(0,len(a)):
b[a[i]]+=1
str1='the cat sat on the hat hat'
list1=str1.split();
list2=str1.split();
count=0;
m=[];
for i in range(len(list1)):
t=list1.pop(0);
print t
for j in range(len(list2)):
if(t==list2[j]):
count=count+1;
print count
m.append(count)
print m
count=0;
#print m