问题描述
Matlab文档表示
a = i;
使用
a = 1i;
稳健性部分很明确，因为可能会有名为i或j 的变量.但是，关于 speed ，我在Matlab 2010b中做了一个简单的测试，得出的结果似乎与主张相矛盾:
The robustness part is clear, as there might be variables called i or j. However, as for speed, I have made a simple test in Matlab 2010b and I obtain results which seem to contradict the claim:
>>clear all
>> a=0; tic, for n=1:1e8, a=i; end, toc
Elapsed time is 3.056741 seconds.
>> a=0; tic, for n=1:1e8, a=1i; end, toc
Elapsed time is 3.205472 seconds.
有什么想法吗?可能是与版本有关的问题吗?
Any ideas? Could it be a version-related issue?
在@TryHard和@horchler发表评论之后，我尝试为变量a分配其他值，结果如下:
After comments by @TryHard and @horchler, I have tried assigning other values to the variable a, with these results:
"i"< "1i"< "1 * i"(趋势"A")
"i" < "1i" < "1*i" (trend "A")
"2i"< "2 * 1i"< "2 * i"(趋势"B")
"2i" < "2*1i" < "2*i" (trend "B")
"1 + 1i"< "1 + i"< "1 + 1 * i"(趋势"A")
"1+1i" < "1+i" < "1+1*i" (trend "A")
"2 + 2i"< "2 + 2 * 1i"< "2 + 2 * i"(趋势"B")
"2+2i" < "2+2*1i" < "2+2*i" (trend "B")
推荐答案
我认为您正在查看一个病理示例.尝试更复杂的操作(在OSX上显示R2012b的结果):
I think you are looking at a pathological example. Try something more complex (results shown for R2012b on OSX):
(重复添加)
>> clear all
>> a=0; tic, for n=1:1e8, a = a + i; end, toc
Elapsed time is 2.217482 seconds. % <-- slower
>> clear all
>> a=0; tic, for n=1:1e8, a = a + 1i; end, toc
Elapsed time is 1.962985 seconds. % <-- faster
(重复乘法)
>> clear all
>> a=0; tic, for n=1:1e8, a = a * i; end, toc
Elapsed time is 2.239134 seconds. % <-- slower
>> clear all
>> a=0; tic, for n=1:1e8, a = a * 1i; end, toc
Elapsed time is 1.998718 seconds. % <-- faster
这篇关于有关在Matlab中将i和j用作变量的更多信息:speed的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！