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
我使用计数器在1行代码中从文本文件单词生成一个frequency . dict
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])
其他回答
#!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()
您可以使用python中提供的内置函数
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])
上面的代码自动删除列表中的重复项，并打印原始列表和没有重复项的列表中每个元素的频率。
一枪双鸟!X维
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'])
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。
计算元素的频率可能最好使用字典:
b = {}
for item in a:
b[item] = b.get(item, 0) + 1
要删除重复项，请使用set:
a = list(set(a))
