表达式'a' or 'z'总是产生'a'。表达式'a' and 'z'总是产生'z'。它不是某种用于在容器中进行查询的DSL，它是一个简单的布尔表达式（并且find被调用其结果）。如果你想说“字符串中是否有'a'或'z'”，你需要做

var1.find('a') != -1 or var.find('z') != -1

对于第二个（字符串中的'a'和'z'）：

var1.find('a') != -1 and var.find('z') != -1

这是因为find方法实际上不支持or和and，它只支持查询字符串。
那么，到底是怎么回事呢？好吧，原来or和and是可以在字符串上执行的运算符。

'a' and 'z' --> 'z'
'a' or 'z'  --> 'a'

现在你已经知道了，你基本上只是像往常一样搜索'a'和'z'。

def count_tokens(text):
   
    #Tokenizes the given text and returns a dictionary with the count of each distinct token.
   
    # First, split the text into individual words
    words = text.split()

    # Next, create an empty dictionary to hold the token counts
    token_counts = {}

    # Loop over the words and count how many times each one appears
    for word in words:
        if word in token_counts:
            token_counts[word] += 1
        else:
            token_counts[word] = 1

    # Finally, return the token counts dictionary
    return token_counts

text = "This is a clock. This is only a clock."
counts = count_tokens(text)
print(counts)

### stopword function
import nltk
from nltk.corpus import stopwords

def count_tokens(text):
   
    #Tokenizes the given text, removes stopwords, and returns a dictionary with the count of each distinct token.

    # First, split the text into individual words
    words = text.split()

    # Next, remove stopwords from the words
    stop_words = set(stopwords.words('english'))
    words = [word for word in words if word.lower() not in stop_words]

    # Next, create an empty dictionary to hold the token counts
    token_counts = {}

    # Loop over the words and count how many times each one appears
    for word in words:
        if word in token_counts:
            token_counts[word] += 1
        else:
            token_counts[word] = 1

    # Finally, return the token counts dictionary
    return token_counts

text = "This is a clock. This is only a clock."
counts = count_tokens(text)
print(counts)