Eric Hansen - 1 year ago 140

Python Question

Let's say I have some DataFrame (*with about 10000 rows in my case, this is just a minimal example*)

`>>> import pandas as pd`

>>> sample_df = pd.DataFrame(

{'col1': list(range(1, 10)), 'col2': list(range(10, 19))})

>>> sample_df

col1 col2

0 1 10

1 2 11

2 3 12

3 4 13

4 5 14

5 6 15

6 7 16

7 8 17

8 9 18

For my purposes, I need to calculate the series represented by

`ln(col_i(n+1) / col_i(n))`

`col_i`

`n`

I know that I can get the

`>>> sample_df.diff()`

col1 col2

0 NaN NaN

1 1 1

2 1 1

3 1 1

4 1 1

5 1 1

6 1 1

7 1 1

8 1 1

Or the percentage change, which is

`(col_i(n+1) - col_i(n))/col_i(n+1)`

`>>> sample_df.pct_change()`

col1 col2

0 NaN NaN

1 1.000000 0.100000

2 0.500000 0.090909

3 0.333333 0.083333

4 0.250000 0.076923

5 0.200000 0.071429

6 0.166667 0.066667

7 0.142857 0.062500

8 0.125000 0.058824

I have just been struggling with a straightforward way to get the direct division of each consecutive column by the previous. Were I to know how to do that even, I could just apply the natural logarithm to every element in the series after the fact.

Currently to solve my problem, I'm resorting to creating another column shifted with row elements down by 1 for each column and then applying the formula between the two columns. It seems messy and sub-optimal to me, though.

Any help would be greatly appreciated!

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Answer Source

just use np.log:

```
np.log(df.col1 / df.col1.shift())
```

you can also use apply as suggested by @nikita but that will be slower.

in addition, if you wanted to do it for the entire dataframe, you could just do:

```
np.log(df / df.shift())
```

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