use*_*006 2 python performance numpy
我制作了一些计算Cronbach Alpha的代码.但是我使用lambda函数并不太好.有没有办法通过使用lambda而不是svar()函数来减少代码并提高效率,并通过使用numpy数组来摆脱一些for循环?
import numpy as np
def svar(X):
n = float(len(X))
svar=(sum([(x-np.mean(X))**2 for x in X]) / n)* n/(n-1.)
return svar
def CronbachAlpha(itemscores):
itemvars = [svar(item) for item in itemscores]
tscores = [0] * len(itemscores[0])
for item in itemscores:
for i in range(len(item)):
tscores[i]+= item[i]
nitems = len(itemscores)
#print "total scores=", tscores, 'number of items=', nitems
Calpha=nitems/(nitems-1.) * (1-sum(itemvars)/ svar(tscores))
return Calpha
###########Test################
itemscores = [[ 4,14,3,3,23,4,52,3,33,3],
[ 5,14,4,3,24,5,55,4,15,3]]
print "Cronbach alpha = ", CronbachAlpha(itemscores)
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def CronbachAlpha(itemscores):
itemscores = numpy.asarray(itemscores)
itemvars = itemscores.var(axis=1, ddof=1)
tscores = itemscores.sum(axis=0)
nitems = len(itemscores)
return nitems / (nitems-1.) * (1 - itemvars.sum() / tscores.var(ddof=1))
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NumPy内置了方差函数.指定ddof=1使用N-1的分母,给出样本方差.还有一个sum内置的.
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