mane - 1 year ago 89

Python Question

[EDIT: Wrong subject/title of post corrected]

I have a dataframe with some columns like this:

`A B C`

0

4

5

6

7

7

6

5

The

Also, I have a list of 8 elements like this:

`List=[2,5,6,8,12,16,26,32] //There are only 8 elements in this list`

If the element in column A is

How can I do this in one go without looping over the whole dataframe?

The resulting dataframe would look like this:

`A B C D`

0 2

4 12

5 16

6 26

7 32

7 32

6 26

5 16

(Note: The dataframe is huge and iteration is the last option option. But I can also arrange the elements in 'List' in any other data structure like dict if necessary)

Answer Source

IIUC, if you make your (unfortunately named) `List`

into an `ndarray`

, you can simply index into it naturally.

```
>>> m = np.arange(16)*10
>>> m[df.A]
array([ 0, 40, 50, 60, 150, 150, 140, 130])
>>> df["D"] = m[df.A]
>>> df
A B C D
0 0 NaN NaN 0
1 4 NaN NaN 40
2 5 NaN NaN 50
3 6 NaN NaN 60
4 15 NaN NaN 150
5 15 NaN NaN 150
6 14 NaN NaN 140
7 13 NaN NaN 130
```

Here I built a new `m`

, but if you use `m = np.asarray(List)`

, the same thing should work: the values in `df.A`

will pick out the appropriate elements of `m`

.

Note that if you're using an old version of `numpy`

, you might have to use `m[df.A.values]`

instead-- in the past, `numpy`

didn't play well with others, and some refactoring in `pandas`

caused some headaches. Things have improved now.