一行列表中的平方和?
问题描述
为了证明我做了某事.这是我的代码总和为三行.
To demonstrate I have done sth. This is my code do sum in three lines.
l=[1,2,3,4,5];
sumOfList=0
for i in l:
sumOfList+=i*i;
print sumOfList
我很好奇我可以只用一行吗?
I am curious can I do it in just one line?
解决方案
怎么样:
sum(map(lambda x:x*x,l))
我们也使用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
我们可以取第一个元素，得到它的平方，然后将它添加到列表的头部，这样我们就可以确保它是平方的.然后我们继续使用reduce.不值得做所有这些工作，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，我尝试比较三种解决方案，将 range 生成的 10000 个数的平方相加，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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