计算Python句子中字母/单词出现次数的代码（abs初学者）

Question

最后的问题

userinput = input('please enter your sentence here: ')
wannacount = input('which character/word you want to count? ')

# lower casing user input as small & capital letters are not the same in python
lower_in = userinput.lower()
lower_wc = wannacount.lower()

count = 0
for word in lower_in:
    if word == lower_wc:
        count = count +1

# note: words won't be counted with this code, only letters
print(wannacount, 'appears', count, 'times in your sentence')

当仅指定一个字符时，这会给出正确的答案。但是当我输入一个单词/>=2个字符时，答案是0。

问题::

Q1。如何制作所需的计算单词和/或字符的程序？

我尝试用“lower_wc”替换“word”，并且计数给出了一些看似随机的数字。例如，“hello world”中的 11 代表字符“o”。

Q2。 11背后的逻辑是什么？

Answer 1

问题1的答案：

以下代码将帮助您计算一个字符串在另一个字符串中出现的次数。在count函数中，参数text是指要搜索的字符串，参数to_be_searched是指要搜索的字符串，如果设置为 True，布尔参数 case_sensitive 会进行区分大小写的匹配；如果设置为 True，布尔参数 overlapped 会搜索重叠匹配。

import regex

text = "bAnana"
to_be_searched = "ana"

def count(text, to_be_searched, case_sensitive = False, overlapped = False):
    if not case_sensitive:
        to_be_searched = to_be_searched.lower()
        text = text.lower()
    matches = regex.findall(to_be_searched, text, overlapped = overlapped)
    return len(matches)
    
count(text, to_be_searched, case_sensitive = False, overlapped = False) # Ans: 1
count(text, to_be_searched, case_sensitive = False, overlapped = True) # Ans: 2
count(text, to_be_searched, case_sensitive = True, overlapped = False) # Ans: 1
count(text, to_be_searched, case_sensitive = True, overlapped = True) # Ans: 1