我怎样才能使我的功能 运行 更快？
How can i make my function run even more faster?
大家好，我正在解决问题 我对我的第一个函数做了很多更改以达到时间限制
但这真的是我最后的想法，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))
它现在给我这次 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 中由于执行时间限制而失败
所以基本上你能指导我使 words_with_given_shape 功能更加优化吗？
*** 编辑 ***：
我忘了告诉问题是它给出了单词列表和单词形状
形状像 [0,1,1,-1] 这意味着
0当量
1 个字符在字母顺序中位于当前字符之后
-1 个字符在字母顺序中位于当前字符之前
你好
它的
[-1, 1, 0, 1]
答案是找到单词列表中形状与 arg 形状相同的所有单词
通过像所示那样重新格式化您的代码并使用 List Comprehensions，您在 10000 次迭代中的速度提高了 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())
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（在这种情况下这是无用的，因为您可以使用提供的形状）！
大家好，我正在解决问题 我对我的第一个函数做了很多更改以达到时间限制
但这真的是我最后的想法，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))
它现在给我这次 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 中由于执行时间限制而失败
所以基本上你能指导我使 words_with_given_shape 功能更加优化吗？
*** 编辑 ***： 我忘了告诉问题是它给出了单词列表和单词形状 形状像 [0,1,1,-1] 这意味着 0当量 1 个字符在字母顺序中位于当前字符之后 -1 个字符在字母顺序中位于当前字符之前
你好 它的 [-1, 1, 0, 1]
答案是找到单词列表中形状与 arg 形状相同的所有单词
通过像所示那样重新格式化您的代码并使用 List Comprehensions，您在 10000 次迭代中的速度提高了 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())
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（在这种情况下这是无用的，因为您可以使用提供的形状）！