numpy:将函数应用于两个numpy数组并返回两个numpy数组 [英] Numpy: apply function to two numpy arrays and return two numpy arrays

问题描述

我有两个输入numpy数组，分别具有一组点的纬度和经度坐标:lats和lons.

I have two input numpy arrays with, respectively, latitude and longitude coordinates of a set of points: lats and lons.

我继承了一个函数，该函数将每个(lat,lon)对转换为一个(E,N)对:

I have inherited a function which converts each (lat,lon) pair into an (E,N) pair:

def convert(lat,lon): #takes two floats as arguments (unit: degrees)
   ...
   computation #Actual function is too long to post
   ...
   return N,E #returns two floats (unit: meters)

我的问题:如何有效地将相同的函数同时应用于两个输入numpy数组?

My question: how could I efficiently apply the same function simultaneously to both input numpy arrays?

我正在考虑修改该函数，以便它返回一个列表:

I was thinking of modifying the function so that it returns a list:

return [N,E] 

以这样的方式:

rows = int(lat.shape[0]) #lat and lon have the same shape
cols = int(lat.shape[1])
easting=numpy.zeros(shape=(rows,cols))
northing=numpy.zeros(shape=(rows,cols))
for i in range(0, rows):
    for j in range(0, cols):
        northing=convert(lon[i][j])[0] #first element of the returned list
        easting=convert(lat[i][j])[1] #second element of the returned list

我尚未对此进行测试，但是通过查看它，我觉得不太舒服.任何见解将不胜感激.

I have not yet tested this, but by looking at it I don't feel very comfortable this will work. Any insights will be much appreciated.

推荐答案

让我们定义一个简单的转换

Let's define a trivial conversion

def convert(lat, lon):
    return lat*np.pi/180, lon*np.pi/180

frompyfunc是将标量"函数应用于数组的一种有用方法.我们甚至可以让它接受2个数组，并返回2个数组(在一个元组中)

frompyfunc is a useful way of applying 'scalar' function to arrays; We can even have it take in 2 arrays, and return 2 arrays (in a tuple)

In [233]: f = np.frompyfunc(convert,2,2)
In [234]: lats=np.linspace(-45,45,5)
In [235]: lons=np.linspace(0,100,5)
In [236]: out = f(lats, lons)
In [237]: out
Out[237]: 
(array([-0.7853981633974483, -0.39269908169872414, 0.0, 0.39269908169872414,
        0.7853981633974483], dtype=object),
 array([0.0, 0.4363323129985824, 0.8726646259971648, 1.3089969389957472,
        1.7453292519943295], dtype=object))

一个功能是它返回一个对象数组，而您可能想要一个浮点数组:

One feature is that it returns an object array, while you probably want a float array:

In [238]: out[0].astype(float)
Out[238]: array([-0.78539816, -0.39269908,  0.        ,  0.39269908,  0.78539816])

或打开包装:

In [239]: rlat, rlon = f(lats, lons)
In [240]: rlat.astype(float)
Out[240]: array([-0.78539816, -0.39269908,  0.        ,  0.39269908,  0.78539816])

frompyfunc确实遍历输入.在其他测试中，它往往比更明确的循环快2倍.在这种情况下，由于它返回一个元组，因此您不必调用它两次即可得到2个结果.

frompyfunc does iterate through the inputs. In other tests it tends to be 2x faster than more explicit loops. And in this case, since it returns a tuple you don't have to call it twice to get 2 results.

按照书面形式，此convert在数组和标量上的工作效果都一样，所以

As written, this convert works just as well with arrays as with scalars, so

In [241]: convert(lats, lons)
Out[241]: 
(array([-0.78539816, -0.39269908,  0.        ,  0.39269908,  0.78539816]),
 array([ 0.        ,  0.43633231,  0.87266463,  1.30899694,  1.74532925]))

这将比在Python中循环的任何版本都要快.

which will be much faster than any version that loops in Python.

因此，为获得真正的速度，您希望convert直接与数组一起使用.但是，如果无法做到这一点，那么frompyfunc是对自己动手"循环的适度改进.

So for real speed you want convert to work directly with arrays. But if it can't do that, then frompyfunc is a modest improvement over do-it-yourself loops.

frompyfunc的另一个优点-如

 f( lats[:,None], lons[None,:])

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