如何使函数运行得更快？

Question

嘿，伙计们，我正在解决一个问题，我对我的第一个函数做了很多修改，以达到时间限制

但这真的是我的最后一个想法，我不知道如何让它比现在更快

from timeit import default_timer as timer
from datetime import timedelta

start = timer()

testList = ['hello']
wordList = ['hello','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe']
def words_with_given_shape(words,shape):
    itemList = 'abcdefghijklmnopqrstuvwxyz'
    def wordShape(word):
        tempS = []
        termil = len(word)-1
        
        for inx,elem in enumerate(word):
            orderAl = itemList.index(elem)
            if inx < termil:
                orderNl = itemList.find(word[inx+1])
                if orderAl > orderNl:
                    tempS.append(-1)
                if orderAl < orderNl:
                    tempS.append(1)
                if orderNl == orderAl:
                    tempS.append(0) 
        return tempS
    def checkWord(words):
        res = []
        for i in words:
            if wordShape(i)==shape:
                res.append(i)
        return res
    return checkWord(words)
print(words_with_given_shape(wordList,  [-1, 1, 0, 1,1,1,-1]))
print(words_with_given_shape(wordList,  [-1, 1, 0, 1]))
print(words_with_given_shape(wordList,  [-1, 1]))
print(words_with_given_shape(wordList,  [-1, 1, 0, 1,1, 0, 1,-1,1]))
print(words_with_given_shape(testList,  [-1, 1, 0, 1]))
end = timer()
print(timedelta(seconds=end-start))

现在给我的时间是0:00:00.001272

但测试人员似乎需要比这更快的速度，因为在测试12时，由于执行时间限制，测试失败

所以基本上你能指导我用给定的形状函数使单词更加优化吗

***编辑***： 我忘了告诉你问题是它给出了单词列表和单词的形状 形状像[0,1,1，-1]，它的意思是 0均衡器 1个字符位于当前字符后面，按字母顺序排列 -1个字符在当前字符之前，按字母顺序排列

你好 它的 [-1,1,0,1]

答案是在单词列表中找到所有形状与形状arg相同的单词

******回答

感谢@Socowi

这解决了我的问题

def words_with_given_shape(words,shape):
    itemList = 'abcdefghijklmnopqrstuvwxyz'
    def wordShape(word):
        tempS = []
        termil = len(word)-1
        if termil == len(shape):
            for inx,elem in enumerate(word):
                orderAl = itemList.index(elem)
                if inx < termil:
                    orderNl = itemList.find(word[inx+1])
                    if orderAl > orderNl:
                        tempS.append(-1)
                    if orderAl < orderNl:
                        tempS.append(1)
                    if orderNl == orderAl:
                        tempS.append(0) 
        return tempS
    def checkWord(words):
        res = []
        for i in words:
            if wordShape(i)==shape:
                res.append(i)
        return res
    return checkWord(words) 

Added single line 

if termil == len(shape):

Answer 1

通过如图所示重新格式化代码并使用列表理解，您在10000次迭代中加快了100毫秒：

• 初始化：

from timeit import default_timer as timer
from datetime import timedelta
start = timer()
testList = ['hello']
wordList = ['hello','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe','helloetttttttttttttttttttttttttwersdfffffffffffffffffffffffffffffffavvvvvvvvvvvvvvvvvvvvvvvvvqwewqe']

• v1:

def checkcond(x, i, word, itemList):
    _, __ = itemList.index(x),  itemList.find(word[i+1])
    return -1 if _ > __ else 1 if _ < __ else 0

def wordShape(word, itemList):
    return [checkcond(x, i, word, itemList) for i, x in enumerate(word[:-1])]

def words_with_given_shape(words,shape):
    itemList = 'abcdefghijklmnopqrstuvwxyz'
    return [x for x in words if wordShape(x, itemList)==shape]

• v2：（在google colab上更快）

def words_with_given_shape(words,shape):
    itemList = 'abcdefghijklmnopqrstuvwxyz'
    def checkcond(x, i, word):
        _, __ = itemList.index(x),  itemList.find(word[i+1])
        return -1 if _ > __ else 1 if _ < __ else 0
    def wordShape(word):
        return [checkcond(x, i, word) for i, x in enumerate(word[:-1])]
    return [x for x in words if wordShape(x)==shape]

• 时间检查：

def timecheck():
    for x in range(10000):
        words_with_given_shape(wordList,  [-1, 1, 0, 1,1,1,-1])
        words_with_given_shape(wordList,  [-1, 1, 0, 1])
        words_with_given_shape(wordList,  [-1, 1])
        words_with_given_shape(wordList,  [-1, 1, 0, 1,1, 0, 1,-1,1])
        words_with_given_shape(testList,  [-1, 1, 0, 1])
    return timedelta(seconds=timer()-start)
print(timecheck())

Answer 2

试试这个：

def words_with_given_shape(words, shape):
    return [word for word in words
            if (len(shape) == len(word) - 1 and
                all(c1 == c2 if s == 0 else (c1 > c2 if s == -1 else c1 < c2)
                    for c1, c2, s in zip(word[:-1], word[1:], shape)))]

它更紧凑（只有一行），并且您不需要生成每个单词的形状（在本例中这是无用的，因为您可以使用提供的形状）