# calculate slope in dataframe

This question is just about calculating the slope at each timestep in a dataframe. There's a lot of extra detail here, that you are welcome to peruse or not, but that one step is all Im looking for.

I have a forecast and an observed dataframe. I am trying to calculate the "interesting" changes in the forecast.

I'd like to try to accomplish that by:

- calculate the best fit of the observed data (ie, linear regression).
- find its slope
- find the difference between the slope and the slope at each moment of the observed data

To do this, I need to generate the slope at each moment in the time series.

- calculate the stddev and mean of that difference
- use that to generate z-scores for the values in the forecast DF.

How do I calculate the slope at each point in the data?

## original

```
from sklearn import linear_model
original = series.copy() # the observations
f = y.copy() # the forecast
app = ' app_2'
original.reset_index(inplace=True)
original['date'] = pd.to_timedelta(original['date'] ).dt.total_seconds().astype(int)
# * calculate the best fit of the observed data (ie, linear regression).
reg = linear_model.LinearRegression()
# * find its slope
reg.fit(original['date'].values.reshape(-1, 1), original[app].values)
slope = reg.coef_
# * find the difference between the slope and the slope at each moment of the observed data
delta = original[app].apply(lambda x: abs(slope - SLOPE_OF(x)))
# * calculate the stddev and mean of that difference
odm = delta.mean()
ods = delta.std(ddof=0)
# * use that to generate z-scores for the values in the forecast DF.
# something like
f['test_delta'] = np.cumsum(f[app]).apply(lambda x: abs(slope - x))
f['z'] = f['test_delta'].apply(lambda x: x - odm / ods)
# from that I might find interesting segments of the forecast:
sig = f.index[f['z'] > 2].tolist()
```

roberto tomás ask on

## 1 Answers calculate slope in dataframe

To "calculate the slope at each point in the data," the simplest is to compute "rise over run" for each adjacent row using

`Series.diff()`

as follows. The resulting Series gives (an estimate of) the instantaneous rate of change (IROC) between the previous and current row.Also, you don't need

`apply`

. Thanks to numpy vectorization,`scalar - array`

behaves as expected:Hope this works. As Wen-Ben commented, it would really help to see actual data and your expected output.

## Peter Leimbigler 1 weeks ago