how to convert this into a column in a new dataframe

Question

well, I am working with a dataframe and I simply want the result that comes out of here to convert it into a new column in a new dataframe. But I do not know how. would you help me? thanks

cont = 12 
for e in df_infl['infl_gen']:
   resultados = (e/df_infl.iloc[cont, 0]) - 1
   resultados = resultados * 100
   cont += 1
if cont >= df_infl.shape[0]:
    print("Ready")
    break

Welcome to stack overflow! Please see [How to make good pandas examples](https://stackoverflow.com/questions/20109391/how-to-make-good-reproducible-pandas-examples) and [edit] your question to include a [mcve] with sample input and expected output so that we can better understand how to help you — G. Anderson, Nov 17 '20 at 19:43
we cant help if we don't know what your data looks like. add `df.to_dict()` output or something — anon01, Nov 17 '20 at 19:48

Raghav Gupta · Answer 1 · 2020-11-17T20:07:07.667

1

Just make a list which has all your data and then simply add a column to DataFrame.

cont = 12 
new_column=[0]*12 #Create a list with 12 values as zero (Insertion has to happen from 12th index)
for e in df_infl['infl_gen']:
   resultados = (e/df_infl.iloc[cont, 0]) - 1
   resultados = resultados * 100
   new_column.append(resultados) #Append your result into list
   cont += 1

   if cont >= df_infl.shape[0]:
       print("Ready")
       break

df_infl['new_column_name']=new_column #Add a new column to your current DataFrame

edited Nov 17 '20 at 20:07

answered Nov 17 '20 at 19:54

Raghav Gupta

454
3
12

1

Are you shure, that would work? I mean, it looks like he is skipping rows (cont=12). So you probably need to insert 12 dummy-values in your list, to make it work, or am I missing something? – jottbe Nov 17 '20 at 20:01
1

edits made! Thank you.. – Raghav Gupta Nov 17 '20 at 20:07
It works! but I guess he wouldn't have to actually loop over the code – jottbe Nov 17 '20 at 20:10

jottbe · Accepted Answer · 2020-11-17T20:31:38.117

If I got your logic right, you can simplify it and make it much faster (because you avoid the loop and can make use of the fast C++ implementations) like this:

df_infl['resultados']= ((df_infl['infl_gen'].shift(12).div(df_infl[name_of_col_0]) - 1) * 100).fillna(0)

For a generated dataframe I get the following output:

    name_col0  infl_gen  new_column_name  resultados
0         127        52         0.000000    0.000000
1         123        68         0.000000    0.000000
2          57       144         0.000000    0.000000
3          57        88         0.000000    0.000000
4         103         4         0.000000    0.000000
5         198       177         0.000000    0.000000
6          79        43         0.000000    0.000000
7         184       183         0.000000    0.000000
8          41        85         0.000000    0.000000
9          15       100         0.000000    0.000000
10         82       105         0.000000    0.000000
11        192       102         0.000000    0.000000
12         74       144       -29.729730  -29.729730
13         53         2        28.301887   28.301887
14         21       132       585.714286  585.714286
15         59        42        49.152542   49.152542
16        161        19       -97.515528  -97.515528
17        132       158        34.090909   34.090909
18        194       110       -77.835052  -77.835052
19        193       171        -5.181347   -5.181347

new_column_name is actually the value calculated by the logic posted by Raghva. So you see both evaluate to the same result.

Yes seems like this would work faster. Time complexity of `for` loop is unbearable! — Raghav Gupta, Nov 17 '20 at 20:42

how to convert this into a column in a new dataframe

2 Answers2