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I need to change the x-axis on this graph from 'samples' to time: enter image description here

I created the graph using this process:

y, _ = librosa.load('sound/data/kea-song.mp3', 48000)
y /= y.max()

# compute the rmse
e = librosa.feature.rmse(y=y)[0]
e -= e.min()#what does this really do, guessing a kind of normalisation?
e /= e.max()
plt.plot(e)
plt.show()

I know the whole len(e)/sr=time formula but I cant actually apply it to the graph. I try this:

plt.plot(len(y)/48000, e)

But I have np.shape issues:

ValueError: x and y must have same first dimension, but have shapes (1,) and (5846,)

How do I do this?

Edit: I tried to do it using both of these (with error's afterwards):

plt.plot(y/48000, e)
ValueError: x and y must have same first dimension, but have shapes (2993006,) and (5846,)

plt.plot(e/48000, e)

linear!

Edit2:

plt.plot(np.arange(len(y))/48000, e)
ValueError: x and y must have same first dimension, but have shapes (2993006,) and (5846,)

plt.plot(np.arange(len(e))/48000, e) ended up with a graph with one vertical line down the middle.

-dear everyone this is what happens when you try do this with little background knowledge of the subject. Do your math homework!

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You're getting an error because the two inputs to plt.plot need to be the same length. Notice that len(y)/48000 is just a number, so python tells us it of length 1.

Try plt.plot(range(len(e))/ 48000, e) **. What this does is change the values of your x axis to correspond to real time. If the sampling rate is $48000 \textrm{Hz}$, this means that $\textrm{sample}=1$ on your current axis now corresponds to 1/48000 of a second in the new axis, as this is the amount of time required to get one sample. Extend this to the whole axis - the time value on the new $x-$axis is simply the amount of time needed to take $y$ samples.

** if range() doesnt work, try numpy.arange() instead

Also, yes, these two lines

e -= e.min()
e /= e.max()

correspond to first shifting the signal so that its minimum value is now at 0, then dividing by the max. This is just a normalization, as you guessed.

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  • $\begingroup$ Hey goldrik cheers for the help. But it doesn't seem to work: I've tried it both using plt.plot(y/48000, e) and plt.plot(e/48000, e). Ill update my main question to explain $\endgroup$ – Finn Maunsell Nov 10 '17 at 10:23
  • $\begingroup$ Ahh, I realized my mistake. We have to specify the new axis here. Recreate the original axis using range (if range() doesnt work, try numpy.arange() instead), which is just a sequence from 0 to len(y). THEN we should do the division to get it in terms of real time. $\endgroup$ – goldrik Nov 10 '17 at 10:32
  • $\begingroup$ Sorry, I meant len(e) here. $\endgroup$ – goldrik Nov 10 '17 at 10:56
  • $\begingroup$ Didn't work again. About to go to bed. Thanks for the help man. $\endgroup$ – Finn Maunsell Nov 10 '17 at 11:17

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