blocks|key|4339798|text|看起来interp2d()的行为是这样的，因为相应的Fortran函数就是这样构思出来的。唯一的解决方法(我能想到的)是对坐标对调用f：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4339799|[f(*p)[0]+for+p+in+zip(x,+y)]|code-block|syntax|javascript|4339800|entityMap^0|3|A|1U|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

It seems that <code>interp2d()</code> behaves they way it does because that is how the corresponding Fortran function was conceived. The only workaround to this (that I can think of) is to call <code>f</code> on pairs of coordinates:

<pre><code>[f(*p)[0] for p in zip(x, y)]
</code></pre>

blocks|key|4339852|text|方法interp2d返回一个对象，该对象的调用方法要求x，y向量是矩形网格的坐标。你不能从返回的数组的对角线得到所需值的原因是，它首先对x，y进行排序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4339853|但是有一个变通方法，我也在Querying+multiple+points+on+a+bivariate+spline+in+the+B-spline+basis中使用过。执行后|4339854|import+scipy.interpolate+as+si
f+=+si.interp2d(x,+y,+z)|code-block|syntax|javascript|4339855|计算f的方法不是调用它，而是将它的tck属性传递给内部bispeu方法，后面是x，y坐标。如下所示：|4339856|print(si.dfitpack.bispeu(f.tck[0],+f.tck[1],+f.tck[2],+f.tck[3],+f.tck[4],+x,+y)[0])|4339857|上面的返回结果与慢速循环相同|4339858|print(np.array([f(xi,+yi)[0]+for+xi,+yi+in+zip(x,+y)]))|4339859|解释|4339860|对象f实际上是一条1阶的B样条曲线。样条参数(节点、系数、阶数)包含在其tck属性中，可以由低阶例程直接使用以达到所需的效果。|4339861|(理想情况下，f的调用方法应该有一个布尔参数grid，我们将该参数设置为False，以便让它知道我们不需要网格求值。遗憾的是，它没有实现。)|4339862|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/46983476^0|2|8|0|D|1W|0|0|0|H|3|R|6|0|0|0|0|0|2|1|10|3|0|7|1|M|4|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1B|8|@]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|H|3|I|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1M|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1N|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1O|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1P|8|@$9|1Q|A|1R|B|C]|$9|1S|A|1T|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1U|8|@$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|10|3|-4|5|6|7|1Z|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

The method <code>interp2d</code> returns an object whose call method expects the x, y vectors to be coordinates of a rectangular grid. And the reason you don't get the desired values from the diagonal of the returned array is that it sorts x, y first. 

But there is a workaround, which I also used in <a href="https://stackoverflow.com/q/46983476">Querying multiple points on a bivariate spline in the B-spline basis</a>. After executing 

<pre><code>import scipy.interpolate as si
f = si.interp2d(x, y, z)
</code></pre>

evaluate f not by calling it, but by passing its <code>tck</code> properties, followed by your x, y coordinates, to internal <code>bispeu</code> method. Like this:

<pre><code>print(si.dfitpack.bispeu(f.tck[0], f.tck[1], f.tck[2], f.tck[3], f.tck[4], x, y)[0])
</code></pre>

The above returns the same as the slow loop

<pre><code>print(np.array([f(xi, yi)[0] for xi, yi in zip(x, y)]))
</code></pre>

<h3>Explanation</h3>

The object <code>f</code> is secretly a B-spline of order 1. The spline parameters (knots, coefficients, order) are contained in its <code>tck</code> property and can be used directly by lower-order routines to the desired effect. 

(Ideally, the call method of <code>f</code> would have a Boolean parameter <code>grid</code> which we'd set to False to let it know we don't want grid evaluation. Alas, it's not implemented.)

blocks|key|4340351|text|根据user6655984的建议，我在另一个thread中发布了以下包装器函数|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4340352|import+scipy.interpolate+as+si
def+interp2d_pairs(*args,**kwargs):
++++"""+Same+interface+as+interp2d+but+the+returned+interpolant+will+evaluate+its+inputs+as+pairs+of+values.
++++"""
++++#+Internal+function,+that+evaluates+pairs+of+values,+output+has+the+same+shape+as+input
++++def+interpolant(x,y,f):
++++++++x,y+=+np.asarray(x),+np.asarray(y)
++++++++return+(si.dfitpack.bispeu(f.tck[0],+f.tck[1],+f.tck[2],+f.tck[3],+f.tck[4],+x.ravel(),+y.ravel())[0]).reshape(x.shape)
++++#+Wrapping+the+scipy+interp2+function+to+call+out+interpolant+instead
++++return+lambda+x,y:+interpolant(x,y,si.interp2d(*args,**kwargs))

#+Create+the+interpolant+(same+interface+as+interp2d)
f+=+interp2d_pairs(X,Y,Z,kind='cubic')
#+Evaluate+the+interpolant+on+each+pairs+of+x+and+y+values
z=f(x,y)|code-block|syntax|javascript|4340353|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/35360756/scipy-interp2d-for-pairs-of-coordinates^0|M|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Based on user6655984's suggestion, I've posted the following wrapper function in another <a href="https://stackoverflow.com/questions/35360756/scipy-interp2d-for-pairs-of-coordinates">thread</a>:

<pre class="lang-py prettyprint-override"><code>import scipy.interpolate as si
def interp2d_pairs(*args,**kwargs):
 """ Same interface as interp2d but the returned interpolant will evaluate its inputs as pairs of values.
 """
 # Internal function, that evaluates pairs of values, output has the same shape as input
 def interpolant(x,y,f):
 x,y = np.asarray(x), np.asarray(y)
 return (si.dfitpack.bispeu(f.tck[0], f.tck[1], f.tck[2], f.tck[3], f.tck[4], x.ravel(), y.ravel())[0]).reshape(x.shape)
 # Wrapping the scipy interp2 function to call out interpolant instead
 return lambda x,y: interpolant(x,y,si.interp2d(*args,**kwargs))

# Create the interpolant (same interface as interp2d)
f = interp2d_pairs(X,Y,Z,kind='cubic')
# Evaluate the interpolant on each pairs of x and y values
z=f(x,y)
</code></pre>

I have data <code>z</code> sampled from a 2D function <code>f</code> at grid points <code>x, y</code>, as in <code>z = f(x, y)</code>.

It is easy to interpolate <code>f</code> with <code>scipy.interp2d</code> via <code>f = interp2d(x, y, z</code>).

However, evaluating <code>f(x, y)</code> returns an entire 2D grid as if I had done

<pre><code>xx, yy = np.meshgrid(x, y)
f(xx, yy)
</code></pre>

The behaviour I want is to simply return the values <code>[f(x[i], y[i]) for i in range(len(x))]</code>, which I believe is the behaviour for pretty much any other method in numpy.

The reason I want this behaviour is that I'm looking for the path traced out along the surface of <code>f</code> over "time" by the pair <code>(t, u(t))</code>.

It is also surprising that <code>np.diag(f(t, u(t)))</code> differs from <code>np.array([f(ti, u(ti)) for ti in t])</code>, so It's not clear to me how to get at the path <code>f(t, u(t))</code> from what is returned via <code>interp2d</code>.

EDIT: About <code>diag</code>, I just thought it seemed we should have <code>np.diag(f(t, u(t))) == np.array([f(ti, u(ti)) for ti in t])</code>, but that isn't the case.

Complete example:

<pre><code>def f(t, u):
 return (t**2) * np.exp((u**2) / (1 + u**2))

x = np.linspace(0, 1, 250)
xx, yy = np.meshgrid(x, x)

z = f(xx, yy)
f = scipy.interpolate.interp2d(x, y, z)

print(f(x, y))
print(np.array([f(xi, yi)[0] for xi, yi in zip(x, y)]))
</code></pre>

I would like the output of both <code>print</code> statements to be the same.

Evaluate the output from scipy 2D interpolation along a curve

我从2D函数f的网格点x, y处采样了数据z，就像在z = f(x, y)中一样。使用scipy.interp2d通过f = interp2d(x, y, z对f进行插值很容易)。但是，计算f(x, y)会返回整个2D网格，就好像我已经这样做了一样xx, yy = np.meshgrid(x, y)f(xx, yy)我...

问评估沿曲线的scipy 2D插值的输出
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