找到最频繁的字符在字符串 [英] Finding the most frequent character in a string
问题描述
我发现这个规划问题,而看一个招聘启事上左右。我认为这是pretty的有趣,作为一个初学者Python程序员我试图解决它。不过,我觉得我的解决方案是相当......凌乱......任何人都可以做任何建议,以优化它或使其更清洁?我知道这是pretty的微不足道,但我很开心写它。注:Python的2.6
的问题:
写假code(或实际code)的函数,它接受一个字符串,并返回出现在最该字符串信。
我的尝试:
输入字符串
高清find_max_letter_count(字):
字母= string.ascii_lowercase
词典= {}
在字母:
词典[来信] = 0
在单词的字母:
词典[来信] + = 1
词典=排序(dictionary.items()
反向=真,
键=拉姆达X:X [1])
有效范围内的位置(0,26):
打印词典[位置]
如果位置= LEN(字典) - 1!
如果词典[位置+ 1] [1];词典[位置] [1]:
打破
find_max_letter_count(HelloWorld的)
输出:
>>>
(L,3)
更新例如:
find_max_letter_count(热气球)
>>>
(L,2)
(O,2)
有很多方法可以做到这一点更短。例如,您可以使用 计数器
类(在Python 2.7或更高版本):
导入收藏
S =的HelloWorld
打印(collections.Counter(S).most_common(1)[0])
如果你没有,你可以做手工理货(2.5或更新版本 defaultdict
):
D = collections.defaultdict(INT)
对于C在S:
D [C] + = 1
打印(排序(d.items(),键=拉姆达X:X [1],反向=真)[0])
说了这么多,有没有什么太可怕的错误与您的实现。
I found this programming problem while looking at a job posting on SO. I thought it was pretty interesting and as a beginner Python programmer I attempted to tackle it. However I feel my solution is quite...messy...can anyone make any suggestions to optimize it or make it cleaner? I know it's pretty trivial, but I had fun writing it. Note: Python 2.6
The problem:
Write pseudo-code (or actual code) for a function that takes in a string and returns the letter that appears the most in that string.
My attempt:
import string
def find_max_letter_count(word):
alphabet = string.ascii_lowercase
dictionary = {}
for letters in alphabet:
dictionary[letters] = 0
for letters in word:
dictionary[letters] += 1
dictionary = sorted(dictionary.items(),
reverse=True,
key=lambda x: x[1])
for position in range(0, 26):
print dictionary[position]
if position != len(dictionary) - 1:
if dictionary[position + 1][1] < dictionary[position][1]:
break
find_max_letter_count("helloworld")
Output:
>>>
('l', 3)
Updated example:
find_max_letter_count("balloon")
>>>
('l', 2)
('o', 2)
There are many ways to do this shorter. For example, you can use the Counter
class (in Python 2.7 or later):
import collections
s = "helloworld"
print(collections.Counter(s).most_common(1)[0])
If you don't have that, you can do the tally manually (2.5 or later has defaultdict
):
d = collections.defaultdict(int)
for c in s:
d[c] += 1
print(sorted(d.items(), key=lambda x: x[1], reverse=True)[0])
Having said that, there's nothing too terribly wrong with your implementation.
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