# How to calculate slope in numpy

If I have an array of 50 elements, how would I calculate a 3 period slope and a 5 period slope? The docs dont add much.....

```
>>> from scipy import stats
>>> import numpy as np
>>> x = np.random.random(10)
>>> y = np.random.random(10)
>>> slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)
```

Would this work?

```
def slope(x, n):
if i<len(x)-n:
slope = stats.linregress(x[i:i+n],y[i:i+n])[0]
return slope
```

but the would the arrays be the same length

@joe:::

```
xx = [2.0 ,4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30]
x = np.asarray(xx, np.float)
s = np.diff(x[::3])/3
window = [1, 0, 0, 0, -1]
window2 = [1, 0, -1]
slope = np.convolve(x, window, mode='same') / (len(window) - 1)
slope2 = np.convolve(x, window2, mode='same') / (len(window2) - 1)
print x
print s
print slope
print slope2
```

Results.....

```
[ 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30.]
[ 2. 2. 2. 2.]
[ 1.5 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. -6. -6.5]
[ 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. -14.]
```

The slope and slope2 are what Im after except the -6, -6.5 and -14 arent the results I am looking for.

this worked.......

```
window = [1, 0, 0, -1]
slope = np.convolve(xx, window, mode='valid') / float(len(window) - 1)
padlength = len(window) -1
slope = np.hstack([np.ones(padlength), slope])
print slope
```

Asked by:

**Catherine305**| Posted: 27-01-2022

# Answer 1

I'm assuming you mean the slope calculated on every 3rd and 5th element so that you have a series of (exact, not least-squares) slopes?

If so, you'd just do something along the lines of:

```
third_period_slope = np.diff(y[::3]) / np.diff(x[::3])
fifth_period_slope = np.diff(y[::5]) / np.diff(x[::5])
```

I'm probably entirely misunderstanding what you mean, though. I've never head the term "3 period slope" before...

If you want more of a "moving window" calculation (so that you have the same number of input elements as output elements), just model it as a convolution with a window of `[-1, 0, 1]`

or `[-1, 0, 0, 0, 1]`

.

E.g.

```
window = [-1, 0, 1]
slope = np.convolve(y, window, mode='same') / np.convolve(x, window, mode='same')
```

Answered by: **Samantha465**| Posted: 28-02-2022

# Answer 2

Just use the subset of the data that contains the points (periods -- I'm assuming you're talking about financial data here) you're interested in:

```
for i in range(len(x)):
if i<len(x)-3:
slope, intercept, r_value, p_value, std_err = stats.linregress(x[i:i+3],y[i:i+3])
if i<len(x)-5:
slope, intercept, r_value, p_value, std_err = stats.linregress(x[i:i+5],y[i:i+5])
```

(This isn't the most efficient approach, btw, if all you want is the slopes, but it's easy.)

Answered by:**Rafael923**| Posted: 28-02-2022

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