Question

我想知道这里是否有人在仍然使用基本字符串方法的同时，有一些浓缩此代码的技巧。该代码要求用户输入其名字，并使用函数和简单的字符串方法显示每个元音首次出现的索引。这是我的代码：

def find_vowel():
    name = input("What is your first name?: ")
    name = name.lower()
    if name.find('a') > 0:
        print "There is an a in your name, first found at index",name.find('a')
    if name.find('e') > 0:
        print "There is an e in your name, first found at index",name.find('e')        
    if name.find('i') > 0:
        print "There is an i in your name, first found at index",name.find('i')  
    if name.find('o') > 0:
        print "There is an o in your name, first found at index",name.find('o')
    if name.find('u') > 0:
        print "There is an u in your name, first found at index",name.find('u')

find_vowel()

是否有任何关于压缩代码的反馈？预先感谢。

Answer 1

而不是为每个元音重复所有这些代码，而是制作一个将其带入变量的循环，如下所示：

def find_vowel():
    name = input("What is your first name?: ")
    name = name.lower()
    for vowel in ['a', 'e', 'i', 'o', 'u']:
        if name.find(vowel) > 0:
            print "There is an %c in your name, first found at index" % vowel,name.find(vowel)

find_vowel()

旁注：您可能不想使用input，而是使用raw_input，因为您仍在使用Python 2。

Answer 2

如果切换到Python 3，其中void addEntry(PQ *pq, void *entry) { assert(pq != NULL); int idx = pq->count; if(pq->count == pq->length) pq = realloc(pq, sizeof(PQ *) * (pq->length * 2)); pq->data[idx] = entry; /* If the last element is greater than its parents, then swap them */ while(idx != 0 && pq->compare(pq->data[PARENT(idx)], pq->data[idx]) > 0) { swap(pq->data[PARENT(idx)], pq->data[idx]); idx = PARENT(idx); } pq->count++; return; }是函数而不是语句，则可以将其压缩为一行，作为列表理解。

print

关于压缩简单python代码的反馈

2 个答案: