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
下面是使用itertools的另一个简洁的替代方案。Groupby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
format: {value: num_of_occurencies}
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
其他回答
对于您的第一个问题,迭代列表并使用字典跟踪元素的存在。
对于你的第二个问题,只需使用集合操作符。
#!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()
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counts = dict.fromkeys(a, 0)
for el in a: counts[el] += 1
print(counts)
# {1: 4, 2: 4, 3: 2, 4: 1, 5: 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。
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