通过另一个函数调用函数时的双输出

Question

count调用该函数find以查看从给定索引开始的单词中可以找到多少次字母(请参阅下面的"代码").

令人困惑的部分:通过使用函数"count",我得到以下程序输出:

可以看出,一些输出是重复的(标记为红色).如果不从发现中删除打印件,如何避免这种情况？有可能还是我被迫删除它(打印)？我知道这两个函数可以变成一个更简单的函数,但我想了解如何使用另一个函数调用函数.

我还必须提到变量计数的值是正确的.唯一的问题是重复的输出.

代码:

def find(word, letter, index):
    start_ind = index
    while index < (len(word)):
        if word[index] == letter:
            print "%s found at index %s" % (letter, index)
            return index

        index += 1

    else:
        print "%s is not found in string '%s' when starting from index %s" % (letter, word, start_ind)
        return -1


def count(word, letter, index):
    count = 0
    while index < len(word):
        if find(word, letter, index) != -1:
            count += 1
            index = find(word, letter, index) + 1

    print "%s is shown %s times in '%s'" % (letter, count, word)

    count("banana", "a", 0)

Answer 1

find()循环中每次迭代有两个调用while:

 if  find(word, letter, index)!= -1:
        count += 1
        index = find(word, letter, index) + 1

每次打印时:

print "%s found at index %s" % (letter,index)

你应该通过计算和存储一次的值来"memoize":find()

 found = find(word, letter, index)
 if  found != -1:
        count += 1
        index = found + 1

这是一个更优雅的问题解决方案:

word = 'banana'
letter = 'a'
occurences = [(index, value) for index, value in enumerate(word) if value == letter]
for occurence in occurences:
    print "Letter ",letter," has been found at index ",index
print "Letter ", letter, " has been found a total of ", len(occurences), " times."