Question

问题＆＃34; Efficiently create a density plot for high-density regions, points for sparse regions＆＃34;要求用NaN替换低密度区域。接受的答案中的相关代码如下：

hh, locx, locy = scipy.histogram2d(xdat, ydat, range=xyrange, bins=bins)
posx = np.digitize(xdat, locx)
posy = np.digitize(ydat, locy)

#select points within the histogram
ind = (posx > 0) & (posx <= bins[0]) & (posy > 0) & (posy <= bins[1])
hhsub = hh[posx[ind] - 1, posy[ind] - 1] # values of the histogram where the points are
xdat1 = xdat[ind][hhsub < thresh] # low density points
ydat1 = ydat[ind][hhsub < thresh]
hh[hh < thresh] = np.nan # fill the areas with low density by NaNs

我发现了像

这样的东西

hh = np.where(hh > thresh, hh, np.nan)

也在工作。在表现结果方面有什么不同？

Answer 1

高级索引（即您的原始方法）效率更高，而结果相同！

我将这两个选项与以下代码进行了比较：

import numpy as np
import time

t1 = time.time()
for i in xrange(1000):
        a = np.random.rand(10)
        a[a>0.5] = np.nan
t2 = time.time()
print 'adv. idx.: ',t2-t1

t1 = time.time()
for i in xrange(1000):
        a = np.random.rand(10)
        a = np.where(a>0.5,np.nan,a)
t2 = time.time()
print 'np.where: ',t2-t1

结果很明显：

adv. idx.:  0.00600004196167
np.where:  0.0339999198914

np.where明显变慢了！结果相同。但是，由于==会产生np.nan == np.nan，因此无法通过False比较进行验证。

差异np.where和这个解决方案？

1 个答案: