Matlab一度被認(rèn)為是最專業(yè)的數(shù)值計(jì)算工具之一，相信許多同學(xué)都或多或少用過這個工具。相比而言，Python作為一種膠水式的語言，其設(shè)計(jì)之初就不是為科學(xué)計(jì)算服務(wù)的。之前也看到許多人在吐槽說用Python去復(fù)現(xiàn)一些計(jì)算過程時經(jīng)常失敗，因此（包括本人）也懷疑過是Python本身數(shù)值精度不夠?qū)е碌摹Ｄ敲碢ython的精度究竟如何，本文就來一探究竟。為了方便，我們就用線性方程組的求解來對比這一事實(shí)。
1、實(shí)驗(yàn)設(shè)計(jì)
本文考慮的線性方程組：Ax=bAx=b A x = b
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生成隨機(jī)系數(shù)矩陣A和常數(shù)項(xiàng)b對數(shù)據(jù)進(jìn)行截?cái)嗍沟脙煞N程序中數(shù)據(jù)完全相等分別用Matlab和Python求解計(jì)算各自誤差值
關(guān)鍵問題一：保證兩種語言處理的數(shù)據(jù)完全一致 這里第二步其實(shí)很重要，由于Matlab默認(rèn)數(shù)據(jù)類型為double，而Python的默認(rèn)類型為float，如果直接粘貼或者導(dǎo)入就有可能在Python中被截?cái)?#xff0c;從而導(dǎo)致在一開始使用的數(shù)據(jù)就存在誤差，使得實(shí)驗(yàn)結(jié)果不可信。
關(guān)鍵問題二：選取合適的指標(biāo)來進(jìn)行對比 第四步主要是計(jì)算兩種程序跑完后均方根誤差：∥Ax?b∥\|Ax-b\| ∥ A x ? b ∥
2、數(shù)據(jù)生成和處理
首先由Matlab隨機(jī)生成一個矩陣和一個向量：
A = rand(5,5)
b = rand(5,1)
然后在Matlab中將數(shù)據(jù)粘至txt文件，再整理成Python的格式粘生成Numpy數(shù)組，同時也將數(shù)據(jù)重新粘回Matlab代碼。這里原始數(shù)據(jù)是這個樣子：
A的值
0.814723686393179 0.0975404049994095 0.157613081677548 0.141886338627215 0.655740699156587
0.905791937075619 0.278498218867048 0.970592781760616 0.421761282626275 0.0357116785741896
0.126986816293506 0.546881519204984 0.957166948242946 0.915735525189067 0.849129305868777
0.913375856139019 0.957506835434298 0.485375648722841 0.792207329559554 0.933993247757551
0.632359246225410 0.964888535199277 0.800280468888800 0.959492426392903 0.678735154857774
b的值
0.757700000000000
0.743100000000000
0.392200000000000
0.655500000000000
0.171200000000000
當(dāng)然也可以采用round函數(shù)一類的方式進(jìn)行處理。
注意 這種操作雖然看似簡單并且有點(diǎn)笨拙，但能很好地保證二者在初始化時得到的結(jié)果是一樣的。
Matlab的操作相對簡單，直接將txt中的內(nèi)容首尾加上[]，再粘回代碼即可。對于Python要稍加處理，最終的代碼是：
import numpy
as np
A
= np
. array
( [ 0.814723686393179 , 0.905791937075619 , 0.126986816293506 , 0.913375856139019 , 0.632359246225410 , 0.0975404049994095 , 0.278498218867048 , 0.546881519204984 , 0.957506835434298 , 0.964888535199277 , 0.157613081677548 , 0.970592781760616 , 0.957166948242946 , 0.485375648722841 , 0.800280468888800 , 0.141886338627215 , 0.421761282626275 , 0.915735525189067 , 0.792207329559554 , 0.959492426392903 , 0.655740699156587 , 0.0357116785741896 , 0.849129305868777 , 0.933993247757551 , 0.678735154857774 ] ) . reshape
( [ 5 , 5 ] ) . Tb
= np
. array
( [ 0.757700000000000 , 0.743100000000000 , 0.392200000000000 , 0.655500000000000 , 0.171200000000000 ] )
注意這里Python的代碼最后要進(jìn)行一下轉(zhuǎn)置，否則與Matlab的數(shù)據(jù)就就不一致。
整理完成看看數(shù)據(jù)和txt中的形狀就完全一致了：
Aarray
( [ [ 0.81472369 , 0.0975404 , 0.15761308 , 0.14188634 , 0.6557407 ] , [ 0.90579194 , 0.27849822 , 0.97059278 , 0.42176128 , 0.03571168 ] , [ 0.12698682 , 0.54688152 , 0.95716695 , 0.91573553 , 0.84912931 ] , [ 0.91337586 , 0.95750684 , 0.48537565 , 0.79220733 , 0.93399325 ] , [ 0.63235925 , 0.96488854 , 0.80028047 , 0.95949243 , 0.67873515 ] ] ) barray
( [ 0.7577 , 0.7431 , 0.3922 , 0.6555 , 0.1712 ] )
這里b變成了一維數(shù)組但不影響計(jì)算。
3、求解方程組
用Matlab求解線性方程組十分簡單，直接一個反除號就可以搞定：
>> x = A \ bx =-0.84093.77013.8572-8.00502.4445
注意這里x跑出來的值只是顯示值，實(shí)際值的位數(shù)是遠(yuǎn)大于這個值的：
-0.840928749143009 3.77006972003583 3.85716025378275 -8.00495386938441 2.44448015270832
用Python求解線性方程組的方式其實(shí)也很多，最簡單的辦法就是直接用scipy庫的linalg模塊中的solve方法：
from scipy
. linalg
import solvex
= solve
( A
, b
) xarray
( [ - 0.84092875 , 3.77006972 , 3.85716025 , - 8.00495387 , 2.44448015 ] )
為了更準(zhǔn)確地對比數(shù)值的精度，我們將x的值用np.savetxt函數(shù)保存至文件再看具體數(shù)字：
- 8.409287491430091910e-01
3.770069720035832184e+00
3.857160253782746739e+00
- 8.004953869384410226e+00
2.444480152708315313e+00
仔細(xì)觀察發(fā)現(xiàn)，Python輸出的數(shù)值位數(shù)要稍多一些，這是因?yàn)槲覀冇肕atlab粘貼的時候它默認(rèn)顯示的位數(shù)不太一樣。
注：這也是為什么我們一定要在最開始進(jìn)行數(shù)據(jù)統(tǒng)一規(guī)范的原因。Matlab在存儲變量時數(shù)據(jù)是保存了完整的位數(shù)的，但即便從變量查看窗口粘出來的數(shù)值也不一定是完整的數(shù)字。
4、精度驗(yàn)證
error = norm(A*x-b)error =1.5801e-15% 查看15位
% 1.580118849929844e-15
np
. sqrt
( np
. sum ( ( A
. dot
( x
) - b
) ** 2 ) ) 1.580118849929844e-15
注意，由于兩個數(shù)的量級都是1e-15，因此這里只需要看它們各自前16位即可。仔細(xì)觀察不難發(fā)現(xiàn)，兩個程序得到的結(jié)果完全一致。
這里我們把有效數(shù)字從第一位到最后一位排列直接放出來對比
1580118849929843973242195223453751011511480618378089335607228349545039236545562744140625
% matlab
1580118849929843973242195223453751011511480618378089335607228349545039236545562744140625
肉眼觀察信不過，全部放入Python字符串試試對不對：
s1
= '1580118849929843973242195223453751011511480618378089335607228349545039236545562744140625'
s2
= '1580118849929843973242195223453751011511480618378089335607228349545039236545562744140625'
判斷一下是否相等：
s1
== s2
True
搞定！
5、結(jié)論
Python(Numpy, Scipy)的數(shù)值精度與Matlab是完全一致的。
其本質(zhì)的原因來自于二者在矩陣計(jì)算都是使用的LAPACK, 詳情可見：
總結(jié)
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