我有一个整数列表(或者甚至可能是字符串),我想在Python中按出现频率排序,例如:
a = [1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5]
这个列表中数字 5
出现了 4 次,4
出现了 3 次。所以排序后的输出列表将是:
result = [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
我尝试使用 a.count()
,但它只给出元素的出现次数。
我想要对它进行排序。有任何想法如何做到这一点吗?谢谢。
from collections import Counter
print [item for items, c in Counter(a).most_common() for item in [items] * c]
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
甚至更好(高效)的实现
from collections import Counter
from itertools import repeat, chain
print list(chain.from_iterable(repeat(i, c) for i,c in Counter(a).most_common()))
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
或者
from collections import Counter
print sorted(a, key=Counter(a).get, reverse=True)
# [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
如果您偏爱原地排序
a.sort(key=Counter(a).get, reverse=True)
使用Python 3.3和内置的sorted函数,以计数为键:
>>> a = [1,1,2,3,3,3,4,4,4,5,5,5,5]
>>> sorted(a,key=a.count)
[2, 1, 1, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5]
>>> sorted(a,key=a.count,reverse=True)
[5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
list.count
会使其非常低效。 - thefourtheyeIn [15]: a = [1,1,2,3,3,3,4,4,4,5,5,5,5]
In [16]: counts = collections.Counter(a)
In [17]: list(itertools.chain.from_iterable([[k for _ in range(counts[k])] for k in sorted(counts, key=counts.__getitem__, reverse=True)]))
Out[17]: [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
或者:
answer = []
for k in sorted(counts, key=counts.__getitem__, reverse=True):
answer.extend([k for _ in range(counts[k])])
当然,[k for _ in range(counts[k])]
可以替换为[k]*counts[k]
。
因此,第17行变成
list(itertools.chain.from_iterable([[k]*counts[k] for k in sorted(counts, key=counts.__getitem__, reverse=True)]))
Counter
根本无法正常工作。 - Ashwini ChaudharyIn [309]: import numpy as np
In [310]: a = [1, 2, 3, 3, 1, 3, 5, 4, 4, 4, 5, 5, 5]
In [311]: vals, counts = np.unique(a, return_counts=True)
In [312]: order = np.argsort(counts)[::-1]
In [313]: np.repeat(vals[order], counts[order])
Out[313]: array([5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2])
tolist()
方法即可:In [314]: np.repeat(vals[order], counts[order]).tolist()
Out[314]: [5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2]
数组中的出现和等大小集合内的出现:
rev=True
arr = [6, 6, 5, 2, 9, 2, 5, 9, 2, 5, 6, 5, 4, 6, 9, 1, 2, 3, 4, 7 ,8 ,8, 8, 2]
print arr
arr.sort(reverse=rev)
ARR = {}
for n in arr:
if n not in ARR:
ARR[n] = 0
ARR[n] += 1
arr=[]
for k,v in sorted(ARR.iteritems(), key=lambda (k,v): (v,k), reverse=rev):
arr.extend([k]*v)
print arr
[6, 6, 5, 2, 9, 2, 5, 9, 2, 5, 6, 5, 4, 6, 9, 1, 2, 3, 4, 7, 8, 8, 8, 2]
[2, 2, 2, 2, 2, 6, 6, 6, 6, 5, 5, 5, 5, 9, 9, 9, 8, 8, 8, 4, 4, 7, 3, 1]
Dart 解决方案
String sortedString = '';
Map map = {};
for (int i = 0; i < s.length; i++) {
map[s[i]] = (map[s[i]] ?? 0) + 1;
// OR
// map.containsKey(s[i])
// ? map.update(s[i], (value) => ++value)
// : map.addAll({s[i]: 1});
}
var sortedByValueMap = Map.fromEntries(
map.entries.toList()..sort((e1, e2) => e1.value.compareTo(e2.value)));
sortedByValueMap.forEach((key, value) {
sortedString += key * value;
});
return sortedString.split('').reversed. Join();
不是很有趣的方式...
a = [1,1,2,3,3,3,4,4,4,5,5,5,5]
from collections import Counter
result = []
for v, times in sorted(Counter(a).iteritems(), key=lambda x: x[1], reverse=True):
result += [v] * times
reduce(lambda a, b: a + [b[0]] * b[1], sorted(Counter(a).iteritems(), key=lambda x: x[1], reverse=True), [])
4
和3
的顺序是否有影响? - thefourtheye