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
简单的解决方法就是用字典。
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]))
其他回答
还有一种方法是使用字典和列表。数数,下面一种幼稚的做法。
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)
Python 2.7+引入了字典理解。从列表中构建字典将获得计数并去除重复项。
>>> 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]
在Python 2.7(或更新版本)中,可以使用集合。计数器:
>>> import collections
>>> a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
>>> counter = collections.Counter(a)
>>> counter
Counter({1: 4, 2: 4, 5: 2, 3: 2, 4: 1})
>>> counter.values()
dict_values([2, 4, 4, 1, 2])
>>> counter.keys()
dict_keys([5, 1, 2, 4, 3])
>>> counter.most_common(3)
[(1, 4), (2, 4), (5, 2)]
>>> dict(counter)
{5: 2, 1: 4, 2: 4, 4: 1, 3: 2}
>>> # Get the counts in order matching the original specification,
>>> # by iterating over keys in sorted order
>>> [counter[x] for x in sorted(counter.keys())]
[4, 4, 2, 1, 2]
如果您使用的是Python 2.6或更老版本,可以在这里下载实现。
在Python 2.7+中,可以使用集合。计数的计数器
>>> 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]
#!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()