Replace commas with points in a dataframe (comma as decimal separator)

If the data is read from a csv with pandas.read_csv and the use of the comma as a decimal separator is consistent throughout the file, what you can do is simply use the decimal parameter that allows you to specify what the character is used as a decimal point:

import io 
import pandas as pd

csv = io.StringIO(
u'''col_1;col_2
7,477;a
7,4848;b
7,4;c
7,5126;d
7,5029;e
5;f''')   




bd_df = pd.read_csv(csv, sep=";", decimal=",")

The use of io.StringIO is just to simulate a .csv and make a reproducible example. In any case, what would be used instead of csv is the path to the file:

bd_df = pd.read_csv("ruta/al/fichero.csv", sep=";", decimal=",")

With what we obtain what is sought:

>>> bd_df
    col_1  col_2
0  7.4770      a
1  7.4848      b
2  7.4000      c
3  7.5126      d
4  7.5029      e
5  5.0000      f


>>> bd_df.dtypes
col_1    float64
col_2     object
dtype: object

If you want to save the csv again using the dot as a decimal separator, it's ok, if you want your columns to have an x number of decimals you can use float_format :

bd_df.to_csv('out.csv', index=False, header=True,
              decimal='.', sep=',', float_format='%.4f')

so we get the following csv file:

  

col_1, col_2
  7.4770, a
  7.4848, b
  7.4000, c
  7.5126, d
  7.5029, e
  5.0000, f

If the above were not possible (for example, it is used in all columns but not in one) you can use pandas.Series.str.replace :

bd_df = pd.read_csv(csv, sep=";")
bd_df['col_1'] = bd_df['col_1'].str.replace(",", ".").astype(float)

  

Note : pandas.DataFrame.convert_objects is marked as "deprecated" from version 0.18, in favor of pandas.DataFrame.to_* (p.e pandas.DataFrame.to_numeric )

FJSevilla

It has been very helpful for me to know how to add the parameters when exporting as you indicated them, with that, plus this command:

comas_por_puntos = [float(x.replace(',','.')) for x in bd_df['col_1']]

the problem was solved.

The first alternative was also tested, but I think there is a missing "u" before: '' 'F; S, because that throws an error that I managed to correct looking for the error and it suggested me to put it like that --- > u '' 'F; S, ... But I could not solve the problem with that solution. I was able to make it export to me as I wanted but only for those indicated in the code you wrote:

csv = io.StringIO(
'''F;S
7,477;a
7,4848;b
7,4;c
7,5126;d
7,5029;e
5;f''')  

Because in the csv I have about 1900 rows and it seemed strange to have to add a letter to it. But in the end the solution appreciated thanks to you. Greetings.