Question

使用Python 3计算列表（字符串）的长度直方图的最佳方法是什么？

程序将计算所有元素的长度，具有相同长度的组元素并返回每个组的大小，与该组中元素的长度配对。

对于以下输入列表：

lis = [ "this" , "is" , "what" , "i" , "have" , "to" , "do" ]

预期结果是：

Letters per word:

3 with 4 letters
3 with 2 letters
1 with 1 letters

是否有内置的容器或功能可以促进这个？

Answer 1

使用collections.Counter和map：

while (true) {
    client = sock.accept();
    inFromClient = new BufferedReader(new  InputStreamReader(client.getInputStream()));
    cIn = inFromClient.readLine();
    System.out.println("" + cln);
    inFromClient.close() //add this
}

Answer 2

您可以使用Counter from collections计算可迭代的任何可清除值

>>> lis = [ "this" , "is" , "what" , "i" , "have" , "to" , "do" ]
>>> from collections import Counter
>>> count = Counter(len(x) for x in lis)
>>> for key in count:
    print(count[key], 'with', key, 'letters')


1 with 1 letters
3 with 2 letters
3 with 4 letters

Answer 3

我会使用 collections.defaultdict ：

import sys
import collections

def processData(data):
    results = collections.defaultdict(list)
    for word in data:
        results[len(word)].append(word)
    return results

if len(sys.argv) > 1:
    data = sys.argv[1:]
    print "Letters per word:\n"
    for k,v in dict(processData(data)).iteritems():
        print "{} with {} letters --> {}".format(len(v), k, ', '.join(v))

<强>输出：

➜ python test.py stack overflow rocks
Letters per word:

1 with 8 letters --> overflow
2 with 5 letters --> stack, rocks

Answer 4

lis = [ "this" , "is" , "what" , "i" , "have" , "to" , "do" ]
a=[]
for item in lis:
    a.append(len(item)) 
b=set(a) 
print "Letters per words are :"
for item in b:
    k=a.count(item)
    print "%d with %d letters"%(k,item)

计算列表中字符串的字符并分组

4 个答案: