How to calculate slope b/w 2 points?

Question

I have a df as follows:

ContextID   EscRF_P2P_Volt_V    StepID  Time_Elapsed
7289972 12.45421    1   0
7289972 12.45421    1   0.055
7289972 12.45421    2   0.156
7289972 12.45421    2   0.487
7289972 12.45421    2   1.477
7289972 12.45421    2   2.477
7289972 13.18681    2   3.477
7289972 12.45421    2   4.487
7289972 12.45421    2   5.993
7289972 12.45421    2   6.545
7289972 12.45421    5   7.983
7289972 12.45421    5   8.993
7289972 13.18681    5   9.993
7289972 13.18681    5   10.393
7289972 12.45421    5   11.993
7289972 12.45421    5   13.093
7289972 12.45421    5   13.384
7289972 12.45421    5   14.388
7289972 12.45421    5   15.386
7289972 12.45421    5   16.386
7289972 12.45421    5   17.396
7289972 12.45421    5   18.406
7289972 12.45421    5   19.396
7289972 11.72161    5   20.396
7289972 12.45421    5   21.396
7289972 12.45421    7   22.386
7289972 12.45421    7   23.456
7289972 13.18681    7   24.404
7289972 12.45421    12  25.443
7289972 13.18681    12  26.443
7289972 11.72161    12  27.443
7289972 12.45421    12  28.453
7289972 13.18681    12  29.443
7289972 12.45421    12  30.443
7289972 12.45421    12  31.443
7289972 12.45421    15  32.472
7289972 27.10623    15  33.444
7289972 27.10623    16  34.443
7289972 22.71062    16  35.443
7289972 22.71062    17  36.443
7289972 622.7106    19  37.503
7289972 622.7106    19  38.513
7289972 622.7106    19  39.503
7289972 622.7106    19  40.503
7289972 622.7106    19  41.503
7289972 622.7106    19  42.503
7289972 622.7106    19  43.503
7289972 622.7106    19  44.503
7289972 622.7106    19  45.532
7289972 622.7106    19  46.502
7289972 622.7106    19  47.501
7289972 622.7106    19  48.501
7289972 622.7106    19  49.501
7289972 622.7106    19  50.501

What I would like to do is calculate the range of Time_Elapsed and split it in 10 parts and calculate the slope of each part, with x being the Time_Elapsed & y being EscRF_P2P_Volt_V column.

I know I can define the slope as:

def slope(x1, y1, x2, y2):
    m = (y2-y1)/(x2-x1)
    return m

But I am not able to implement it correctly.

Any suggestions as to how can it be done?

Output:

The output for the first interval between 0-5 must be something like this:

slope = (12.45421-12.45421)/(5-0)

For the second interval between 5-10

slope = (13.18681-12.45421)/(10-5)

and so on...

If there is no exact value in Time_Elapsed , like there is no 10 , so in that case we take the EscRF_P2P_Volt_V value at 9.993

Answer 1

You could add another column with 10 Group IDs, then use groupby and calculate the last minus first of the EscRF_P2P_Volt_V column in each Group divided by the last minus first of the Time_Elapsed column:

df['grpNo'] = df.Time_Elapsed // 5.0502

a more general approach for the grpNo calculation if you have a normal counting index from 0... n-1 (but you didn't post that):

df['grpNo'] = df.index.values // (len(df)/10)

But note that they do not result in the same grouping, so the results of the slopes vary, too. It's up to you to implement the grouping you'd like to use...

grpd = df.groupby('grpNo')

(grpd.EscRF_P2P_Volt_V.last() - grpd.EscRF_P2P_Volt_V.first()) / (grpd.Time_Elapsed.last() - grpd.Time_Elapsed.first())

# grpNo
# 0.0      0.000000
# 1.0      0.183150
# 2.0     -0.183379
# 3.0      0.000000
# 4.0      0.365569
# 5.0      0.183150
# 6.0      3.663005
# 7.0    147.783246
# 8.0      0.000000
# 9.0      0.000000
# dtype: float64

result with index based grouping

# grpNo
# 0.0      0.000000
# 1.0     -0.162583
# 2.0      0.000000
# 3.0      0.000000
# 4.0      0.146286
# 5.0      0.183150
# 6.0      3.663005
# 7.0    118.577071
# 8.0      0.000000
# 9.0      0.000000
# dtype: float64

comparison of the group sizes of the different groupings here:

   Index based  Time based
0  6            8
1  5            5
2  6            5
3  5            5
4  6            5
5  5            5
6  5            5
7  6            5
8  5            5
9  5            6