Python 检测列表元素是否为多字

这段代码可以正常工作，但我不能100%确定它是如何工作的，因为它是在我借的一本Python书中。我不明白程序是如何检查某个东西是否是多字的。还有，星号线是什么意思

places= ["Home","In-n Out Burger", "John's house", "Santa Monica Pier", "Staples center",  "LA Dodgers stadium", "Home"]
def placesCount(places):
    multi_word = 0
    count = 0
    **while True:
        place = places[count]**
        if place == 'LA Dodgers stadium':
            break
        **if ' ' in place:**
            multi_word += 1
        count += 1
    return count + 1, multi_word + 1

placesCount(places)

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该方法检查列表

places

中的字符串是否有空格，它认为一个多字

如果列表

places

包含字符串

LA Dodgers stadium

，则该方法将返回字符串的位置，以及在此之前找到的多个单词的计数

这里有一个提示：当您将

['LA Dodgers stadium']

传递给函数时会发生什么？它返回正确的数字吗

def placesCount(places):
    multi_word = 0 # a count of how many multiple words were found
    count = 0 # an initializer (not needed in Python)
    while True: # start a while loop
        place = places[count] # get the object from the places list
                              # at position count
        if place == 'LA Dodgers stadium':
            # quit the loop if the current place equals 'LA Dodgers stadium' 
            break
        if ' ' in place:
            # If the current string from the places list
            # (which is stored pointed to by the name place)
            # contains a space, add one to the value of multi_word
            multi_word += 1
        # Add one to count, so the loop will pick the next object from the list
        count += 1
    # return a tuple, the first is how many words in the list
    # and the second item is how many multiple words (words with spaces)
    return count + 1, multi_word + 1

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if''in place:#如果in place字符串包含空格，请在值中添加一个空格。我不明白它是如何检测是否有空格的。如果没有使用FIND方法，如何查找空格。我需要在不使用查找方法的情况下完成我的程序。你在说什么查找方法？