Codemon - 1 year ago 131

Python Question

I have been trying to make a plot smoother, like it is done here, but my Xs are datetime objects that are not compatible with linspace..

I convert the Xs to matplotlib dates:

`Xnew = matplotlib.dates.date2num(X)`

X_smooth = np.linspace(Xnew.min(), Xnew.max(), 10)

Y_smooth = spline(Xnew, Y, X_smooth)

But then I get an empty plot, as my Y_smooth is

`[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]`

for some unknown reason.

How can I make this work ?

Here's what I get when I print the variables, i see nothing abnormal :

`X : [datetime.date(2016, 7, 31), datetime.date(2016, 7, 30), datetime.date(2016, 7, 29)]`

X new: [ 736176. 736175. 736174.]

X new max: 736176.0

X new min: 736174.0

XSMOOTH [ 736174. 736174.22222222 736174.44444444 736174.66666667

736174.88888889 736175.11111111 736175.33333333 736175.55555556

736175.77777778 736176. ]

Y [711.74, 730.0, 698.0]

YSMOOTH [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

Your `X`

values are reversed.

```
>>> from scipy.interpolate import spline
>>> import numpy as np
>>> spline(
... [736176.0, 736175.0, 736174.0], # <-- your original X is decreasing
... [711.74, 730.0, 698.0],
... np.linspace(736174.0, 736176.0, 10) # <-- this is increasing
... )
array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
```

reverse `X`

and `Y`

first and it works

```
>>> spline(
... [736174.0, 736175.0, 736176.0], # <-- reverse order of X
... [698.0, 730.0, 711.74], # <-- reverse order of Y also
... np.linspace(736174.0, 736176.0, 10) # <-- to match new order
... )
array([ 698. , 262.18297973, 159.33767533, 293.62017489,
569.18656683, 890.19293934, 1160.79538066, 1285.149979 ,
1167.41282274, 711.74 ])
```

cricket_007 suggestion might be better than using spline, as it seems that spline is poorly undocumented and may not be supported. For example, it is not listed on the `scipy.interpolate`

main page or on the interploation tutorial. However `interp1d`

is well documented and explained in detail. However you don't have enough points for a cubic spline, and a quadratic spline is poorly fit.

```
f = interp1d([736176.0, 736175.0, 736174.0], [711.74, 730.0, 698.0], kind='quadratic')
f(np.linspace(736176.0, 736174.0, 10))
array([ 711.74 , 720.14123457, 726.06049383, 729.49777778,
730.45308642, 728.92641975, 724.91777778, 718.4271605 ,
709.4545679 , 698. ])
```

There is also a numpy equivalent that can be used once through although it may also have the x monotonically increasing condition.

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**