计算字符串中出现的次数

Question

我需要返回一个字典，计算预定列表中每个字母出现的次数。问题是我需要将大写和小写字母都计算为相同，所以我不能使用 .lower 或 .upper。

因此，例如，如果“t”是要搜索的字母，“这是一个 Python 字符串”应该返回 {'t':3}。

这是我目前所拥有的......

def countLetters(fullText, letters):
    countDict = {i:0 for i in letters}
    lowerString = fullText.lower()
    for i in lowerString:
        if i in letters:
            countDict[i] += 1

    return countDict

'letters' 是条件，fullText 是我要搜索的字符串。

这里明显的问题是，如果测试是“T”而不是“t”，我的代码将不会返回任何内容抱歉我的术语中有任何错误，我对此很陌生。任何帮助将不胜感激！

Answer 1

要忽略大小写，您需要输入 = 输入 = input.lower () . 使用列表操作列出输入文本的所有字符。 如果您扫描空格字符，它也可以用作单词计数器。

input = "Batuq batuq BatuQ" # Reads all inputs up to the EOF character

input = input.replace('-',' ')#Replace (-, + .etc) expressions with a space character.
input = input.replace('.','')
input = input.replace(',','')
input = input.replace("`",'')
input = input.replace("'",'')

#input= input.split(' ') #if you use it, it will sort by the most repetitive words
dictionary = dict()
count = 0
for word in input:
    dictionary[word] = input.count(word)
    
print(dictionary)

#Writes the 5 most repetitive characters
for k in sorted(dictionary,key=dictionary.get,reverse=True)[:5]:
    print(k,dictionary[k])

Answer 2

这样的东西能同时处理区分大小写的字母计数和不区分大小写的计数吗？

from typing import List

def count_letters(
    input_str: str,
    letters: List[str],
    count_case_sensitive: bool=True
):
    """
    count_letters consumes a list of letters and an input string
    and returns a dictionary of counts by letter. 
    """
    if count_case_sensitive is False:
        input_str = input_str.lower()
        letters = list(set(map(lambda x: x.lower(), letters)))
    # dict comprehension - build your dict in one line
    # Tutorial on dict comprehensions: https://www.datacamp.com/community/tutorials/python-dictionary-comprehension
    counts = {letter: input_str.count(letter) for letter in letters}
 
    return counts

# define function inputs    
letters = ['t', 'a', 'z', 'T']
string = 'this is an example with sTrings and zebras and Zoos'

# case sensitive
count_letters(
    string,
    letters,
    count_case_sensitive=True
)
# {'t': 2, 'a': 5, 'z': 1, 'T': 1}

# not case sensitive
count_letters(
    string,
    letters,
    count_case_sensitive=False
)
# {'a': 5, 'z': 2, 't': 3} # notice input T is now just t in dictionary of counts

Answer 3

试试看 - 像这样：

def count_letters(fullText, letters):
    countDict = {i: 0 for i in letters}
    lowerString = fullText.lower()
    for i in lowerString:
        if i in letters:
            countDict[i] += 1
    return countDict

test = "This is a Python string."
print(count_letters(test, 't')) #Output: 3

Answer 4

您正在循环错误的字符串。您需要循环遍历lowerString，而不是fullString，因此在计数时忽略大小写。

if i in countDict 也比if i in letter 更有效。

def countLetters(fullText, letters):
    countDict = {i.lower():0 for i in letters}
    lowerString = fullText.lower()
    for i in lowerString:
        if i in countDict:
            countDict[i] += 1

    return countDict

Answer 5

你可以做的就是简单地复制大写和小写的字典，如下所示：

def countLetters(fullText, letters):
    countDict = {}
    for i in letters:
        countDict[i.upper()]=0
        countDict[i.lower()]=0
    lowerString = fullText.lower()
    letters = letters.lower()
    for i in lowerString:
        if i in letters:
            countDict[i] += 1
            if (i!=i.upper()):
                countDict[i.upper()] +=1

    return countDict
print(countLetters("This is a Python string", "TxZY"))

现在您还可以做的一些事情是遍历原始字符串并将countDict[i] += 1 更改为countDict[i.lower()] +=1

Answer 6

使用集合模块中的计数器

from collections import Counter
input = "Batuq batuq BatuQ"
bow=input.split(' ')
results=Counter(bow)
print(results)

输出： Counter({'Batuq': 1, 'batuq': 1, 'BatuQ': 1})