给定一个无序的值列表,比如
a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
我怎样才能得到出现在列表中的每个值的频率,就像这样?
# `a` has 4 instances of `1`, 4 of `2`, 2 of `3`, 1 of `4,` 2 of `5`
b = [4, 4, 2, 1, 2] # expected output
给定一个无序的值列表,比如
a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
我怎样才能得到出现在列表中的每个值的频率,就像这样?
# `a` has 4 instances of `1`, 4 of `2`, 2 of `3`, 1 of `4,` 2 of `5`
b = [4, 4, 2, 1, 2] # expected output
当前回答
通过遍历列表并计算它们,手动计算出现的数量,使用collections.defaultdict跟踪到目前为止看到的内容:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1
其他回答
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的数量。
所以如果你想要得到一个数字数组,你应该使用这个:
np.unique(a, return_counts=True)[1]
还有另一种不使用集合的算法解决方案:
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
对于一个无序列表,你应该使用:
[a.count(el) for el in set(a)]
输出为
[4, 4, 2, 1, 2]
#!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()
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