# Longest chain of points satisfying given condition

## Question:

I have a graph of discrete set of points.

```
y x
0 1.000000 1000.000000
1 0.999415 1000.000287
2 0.999420 1000.000358
3 0.999376 1000.000609
4 0.999239 1000.000788
5 0.999011 1000.000967
6 1.000389 1000.001433
7 0.999871 1000.001756
8 0.995070 1000.002723
9 0.996683 1000.003404
```

I want to determine the longest chain of consecutive points where the slope of the line connecting `i-1`

to `i`

remains within a given range `epsilon = 0.4`

.

```
def tangent(df, pt1, pt2):
y = df.iloc[pt2]['y'] - df.iloc[pt1]['y']
x = df.iloc[pt2]['x'] - df.iloc[pt1]['x']
return x/y
```

The data has been normalized to scale the tangent results.

```
index = 1
while index < df.shape[0]:
if abs(math.tan(tangent(df,index-1,index) * math.pi)) < epsilon:
print("result:",index)
index += 1
```

The snippet is the draft to detect all such points.

## Answers:

You can simplify the code using pandas methods which apply to whole column (Series):

```
import numpy as np
...
# equivalent of your `tangent` function
# df['x'].diff() will return a column where every row is
# actual rows difference with previous one
df['tang'] = df['x'].diff()/df['y'].diff()
# np.tan will calculate the tan of whole column values at once
matches = np.tan(df['tang']) < epsilon
# Get longest chain
(~matches).cumsum()[matches].value_counts().max()
```

More info: