计算字母的使用次数

Question

因此，我不仅需要计算单词中字母的出现频率，还需要吐出未使用的字母。

#Dictionary and lists etc...
alphabetList = ["a","b","c","d","e","f","g","h","i","j","k",\
    "l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
#Dictionary and lists etc...


def Checksums():
    for i in alphabetList:

    # Get everything ready to edit
        character = i
        myString = Phrase
        myList = []
    #DONE GATHERING RESOURCE

        for i in myString:
            myList.append(i)

        myList = myList.sort()
        print myList

        count = 0
        for i in myList:
            if i == character:
                count = count +1
            else:
                continue
        print("There are ", count, " occurrences of ", character )


#input
Phrase = str(raw_input("Enter a Phrase: "))

Phrase = Phrase.lower()
# print (Phrase)
Phrase = Phrase.replace(" ", "")
# print Phrase

Checksums()

请原谅我对python的了解，语法与java不同，有时会令人沮丧

输入的示例可能是：aaA cC D并且返回将是

“字母“ a”出现了3次”“出现了字母'c'的'2'个出现”“只有1个'd'”“其余字母未使用”bË等等...

我觉得这可能与我不了解的语法/逻辑有关。例如，我尝试使用字母列表并循环浏览，但是我得到了

TypeError：'NoneType'对象不可迭代

Answer 1

用您的功能中的myList = myList.sort()替换myList.sort()。

Answer 2

使用Python的Counter和set确定缺少的键，结果为：

import string
from collections import Counter


def check(phrase):
    phrase = phrase.lower().replace(" ", "")
    phrase = Counter(phrase)

    for counts in phrase.items():
        print ("There were %s occurrences of the letter %s" % counts[::-1])

    missingCharacter = set(string.ascii_lowercase) - set(phrase.keys())
    print("The remaining letters are unused: %s" % ','.join(missingCharacter))


check('aaA cC D')

输出：

There were 3 occurrences of the letter a
There were 2 occurrences of the letter c
There were 1 occurrences of the letter d
The remaining letters are unused: b,e,g,f,i,h,k,j,m,l,o,n,q,p,s,r,u,t,w,v,y,x,z

Answer 3

一些简化代码的方法

首先，要计算字符串中每个字母的出现次数，请使用字典：

d = {};
for c in myString:
     d[c] = (d[c]+1) if (c in d) else 1

然后打印每个字母的统计信息：

# used letters:
for k in d:
    print("There are ", d[k], " occurrences of ", k)


#unused letters
for o in range(ord('a'),ord('z')):
   k = chr(o)
   if not (k in d):
       print("No occurrence of ", k, " in string")

或合并：

for o in range(ord('a'),ord('z')):
   k = chr(o)
   if not (k in d):
       print("No occurrence of ", k, " in string")
   else:
       print("There are ", d[k], " occurrences of ", k)

Answer 4

您可以尝试使用此

alphabetList = ["a","b","c","d","e","f","g","h","i","j","k",\
"l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
def occurence_counter(phrase):
    unused_letters = ""
    for alphabet in alphabetList:
        count = phrase.count(alphabet)
        if count > 1:
            print("There are ", count, " occurrences of ", alphabet )
        else:
            unused_letters = unused_letters + alphabet
    print("The remaining letters are unused" + unused_letters)
Phrase = str(input("Enter a Phrase: "))

Phrase = Phrase.lower()
occurence_counter(Phrase)

Answer 5

由于嵌套循环，请勿将所有的'i'用作变量