我就廢話不多說了，大家還是直接看代碼吧~

import timeimport mathimport numpy as npdef timeit1(): s = time.time() for i in range(750000): z=i**.5 print ('Took %f seconds' % (time.time() - s))def timeit2(arg=math.sqrt): s = time.time() for i in range(750000): z=arg(i) print ('Took %f seconds' % (time.time() - s))def timeit3(arg=np.sqrt): s = time.time() for i in range(750000): z=arg(i) print ('Took %f seconds' % (time.time() - s))def timeit4(): s = time.time() for i in range(750000): z=math.pow(i,.5) print ('Took %f seconds' % (time.time() - s))timeit1()timeit2()timeit3()timeit4()

Took 0.152364 secondsTook 0.061580 secondsTook 1.016529 secondsTook 0.215403 seconds

補充：Python筆記-開根號的幾種方式

使用Python中的自帶庫math、自帶函數(shù)pow和自帶庫cmath來對數(shù)字進(jìn)行開根號運算

使用：math.sqrt(數(shù)字)

import mathn = int(input(’數(shù)字:’))x = math.sqrt(n)print(x)print(type(x)) #開根號后的類型為float方法二

使用：pow(數(shù)字，次方)

n = int(input(’數(shù)字:’))x = pow(n,0.5)print(x)print(type(x)) #開根號后的類型為float方法三

使用：cmath.sqrt(數(shù)字)

該方法多用于復(fù)數(shù)、負(fù)數(shù)的開方運算

import cmathn = int(input(’數(shù)字: ’))x = cmath.sqrt(n)print(x)print(type(x)) #類型為complex

以上為個人經(jīng)驗，希望能給大家一個參考，也希望大家多多支持好吧啦網(wǎng)。如有錯誤或未考慮完全的地方，望不吝賜教。