在long vector，python中获得最小值索引的有效方法 [英] Efficient way to get index of minimum value in long vector, python

问题描述

我有一长串的经度值（len（Lon）= 420481）和另一个纬度值。我想找到经度最小值的相应纬度。

I have a long list of longitude values (len(Lon) = 420481), and another one of latitude values. I want to find the corresponding latitude to the minimum of the longitude.

我试过：

SE_Lat = [Lat[x] for x,y in enumerate(Lon) if y == min(Lon)]

但这需要很长时间才能完成。

but this takes ages to finish.

有没有人知道更有效的方式？

Does anyone know a more efficient way?

也许你也有这样的建议：
我现在尝试找到与新经度最接近的相应纬度，这不在原始经度向量中。我试过这个：

Maybe you also have a suggestions for this: I now try to find the closest corresponding latitude to a new longitude, which is not in the original longitude vector. I tried this:

minDiff = [min(abs(x - lon_new) for x in lons)] # not very quick, but works
[(lat,lon) for lat,lon in izip(lats,lons) if abs(lon-lon_new)==minDiff]

最后一行抛出错误，因为有多个匹配。我现在不知道如何只找到一个值，让我们说第一个。非常感谢任何帮助！

The last line throws an error, because there are multiple matches. I don't know at the moment how to find only one value, lets say the first. Any help is greatly appreciated!

推荐答案

我可以推荐numpy吗？

May I recommend numpy?

import numpy
nplats = numpy.array(lats)
nplons = numpy.array(lons)

# this part is 20x faster than using the built-in python functions
index = numpy.argmin(nplats)

print nplats[index], nplons[index]

这比min（izip（））解决方案快（使用420481随机创建的记录时使用我的设置约20倍），当然你会需要将您的数据值存储在numpy中以利用此加速。

this is way faster than the min(izip()) solution (~20x using my setup when using 420481 randomly created records), although of course you'd need to store your data values in numpy to take advantage of this speed-up.

这篇关于在long vector，python中获得最小值索引的有效方法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IT屋！