一行中的列表中的平方和? [英] sum of squares in a list in one line?

问题描述

为了证明我已经做了某事. 这是我的代码，总分为三行.

To demonstrate I have done sth. This is my code do sum in three lines.

l=[1,2,3,4,5];
sum=0

for i in l:
    sum+=i*i;
print sum

我很好奇我可以只一行吗?

I am curious can I do it in just one line?

推荐答案

关于:

sum(map(lambda x:x*x,l))

我们还使用reduce:

print reduce(lambda x,y: x+y*y,l) # as pointed by @espang reduce(lambda x,y: x+y*y,l) is only ok, when the first value is 1 (because 1*1 == 1). The first value is not squared

我们可以采用第一个元素，获取其平方，然后将其添加到列表的开头，以确保它是平方的.然后我们继续使用reduce.所有的工作都不值得，因为我们有更好的选择.

We can take the first element, get its square, then add it to the head of the list so we can make sure that it's squared. Then we continue using reduce. It isn't worth all that work, as we have better alternatives.

reduce(lambda x,y: x+y*y,[l[:1][0]**2]+l[1:])

出于好奇，我试图比较这三种解决方案，以求和由range生成的10000数字的平方，并计算每个操作的执行时间.

Just out of curiosity, I tried to compare the three solutions to sum the squares of 10000 numbers generated by range, and compute the execution time of every operation.

l=range(10000) 
from datetime import datetime
start_time = datetime.now()
print reduce(lambda x,y: x+y*y,l)
print('using Reduce numbers: {}'.format(datetime.now() - start_time))

from datetime import datetime
start_time = datetime.now()
print sum(map(lambda x:x*x,l))
print('Sum after map square operation: {}'.format(datetime.now() - start_time))

from datetime import datetime
start_time = datetime.now()
print sum( i*i for i in l)
print('using list comprehension to sum: {}'.format(datetime.now() - start_time))

输出:

使用list comprehension更快

333283335000
using Reduce numbers: 0:00:00.003371
333283335000
Sum after map square operation: 0:00:00.002044
333283335000
using list comprehension to sum: 0:00:00.000916

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