how to pass all the column from dataframe into different user defined function

Question

How to pass all the column from dataframe into different user defined function

here is mine dataframe look like

    data = [['tom', 10, 9876765143, 'SUN 1023'], ['nick', 15, 98767654312, 'SUN 1023'], ['juli', 14, 98769876541, 'SUN 1023']]    
    df = pd.DataFrame(data, columns = ['Name', 'Age', 'Number', 'Address'])         
    df

here are function right now i am just showing one function

        def number(inp):
            import re
            regex = r'^\s*(?:\+?(\d{1,3}))?[-. (]*(\d{3})[-. )]*(\d{3})[-. ]*(\d{4})(?: *x(\d+))?\s*$'
            inp = inp.replace(regex, 'XXXXXXX')
            print (inp)

  number(df.Number)

It throws error

     ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

similarly i have multiple function i just want to pass each column from dataframe with the value associated within the column of dataframe into each user defined function like i have a function number similarly i have other function def new() def beg() one by one. Is there any way to solve that problem

I run your code and see no error, the result is: 0 9876765143 1 98767654312 2 98769876541 — Abdirahman
– Abdirahman, Commented Nov 12, 2019 at 19:23
0 1 and 2 are indexes, 0:9876765143 1:98767654312 2:98769876541 — Abdirahman
– Abdirahman, Commented Nov 12, 2019 at 19:27
i want to use it in a function and i have several other regex too even i have to apply multiple regex to a single column. I just want to pass each column into all the user defined function one by one is it possible? — Parth Tiwari
– Parth Tiwari, Commented Nov 12, 2019 at 19:33
I am not sure if i fully understand what you want, you have more than one function and every function is designed to apply regrex to specific column of the dataframe,is that right? — Abdirahman
– Abdirahman, Commented Nov 12, 2019 at 19:51

Alexander · Accepted Answer · 2019-11-12 19:56:24Z

2

Why even use a function? You can directly apply the regex pattern to the relevant column(s).

regex = r'^\s*(?:\+?(\d{1,3}))?[-. (]*(\d{3})[-. )]*(\d{3})[-. ]*(\d{4})(?: *x(\d+))?\s*$'

>>> df['Number'].astype(str).str.replace(regex, 'XXXXXXX')
0    XXXXXXX
1    XXXXXXX
2    XXXXXXX
Name: Number, dtype: object

If you need a function:

import random

def phone_number(series):
    regex = r'^\s*(?:\+?(\d{1,3}))?[-. (]*(\d{3})[-. )]*(\d{3})[-. ]*(\d{4})(?: *x(\d+))?\s*$'
    return series.str.replace(regex, 'X' * random.randrange(3, 8))

>>> phone_number(df['Number'].astype(str))
0    XXXXXXX
1    XXXXXXX
2    XXXXXXX
Name: Number, dtype: object

edited Nov 12, 2019 at 19:56

answered Nov 12, 2019 at 19:16

Alexander

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6 Comments

Parth Tiwari Over a year ago

Hey @Alexander i want to use it in a function and i have several other regex too even i have to apply multiple regex to a single column. I just want to pass each column into all the user defined function one by one is it possible?

Parth Tiwari Over a year ago

How to iterate all such function to each and every column in dataframe

Parth Tiwari Over a year ago

Can i generate random length 'XXXXX'

Alexander Over a year ago

To apply it to all columns in the dataframe: {col: phone_number(df[col].astype(str)) for col in df}. For specific columns, just replace df with a list of the relevant columns. You could generate random length XXXXX, but that sounds like a separate question.

Parth Tiwari Over a year ago

As i think i just need to use randint to generate random lenghth 'XXXXX' if you could answer me that will be become easy for me thanks anyway

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