根据同一行的其他列中的值将函数应用于dataframe列元素？

Question

I have a dataframe:

df = pd.DataFrame(
    {'number': ['10', '20' , '30', '40'], 'condition': ['A', 'B', 'A', 'B']})

df = 
    number    condition
0    10         A
1    20         B
2    30         A
3    40         B

I want to apply a function to each element within the number column, as follows:

 df['number'] = df['number'].apply(lambda x: func(x))

BUT, even though I apply the function to the number column, I want the function to also make reference to the condition column ie in pseudo code:

func(n):
    #if the value in corresponding condition column is equal to some set of values:
        # do some stuff to n using the value in condition
        # return new value for n

For a single number, and an example function I would write:

number = 10
condition = A
def func(num, condition):
    if condition == A:
        return num*3
    if condition == B:
        return num*4

func(number,condition) = 15

How can I incorporate the same function to my apply statement written above? ie making reference to the value within the condition column, while acting on the value within the number column?

Note: I have read through the docs on np.where() , pandas.loc() and pandas.index() but I just cannot figure out how to put it into practice.

I am struggling with the syntax for referencing the other column from within the function, as I need access to both the values in the number and condition column.

As such, my expected output is:

df = 
    number    condition
0    30         A
1    80         B
2    90         A
3    160         B

UPDATE: The above was far too vague. Please see the following:

df1 = pd.DataFrame({'Entries':['man','guy','boy','girl'],'Conflict':['Yes','Yes','Yes','No']})


    Entries    Conflict
0    "man"    "Yes"
1    "guy"    "Yes"
2    "boy"    "Yes"
3    "girl"   "No

def funcA(d):
    d = d + 'aaa'
    return d
def funcB(d):
    d = d + 'bbb'
    return d

df1['Entries'] = np.where(df1['Conflict'] == 'Yes', funcA, funcB)

Output:
{'Conflict': ['Yes', 'Yes', 'Yes', 'Np'],
 'Entries': array(<function funcB at 0x7f4acbc5a500>, dtype=object)}

How can I apply the above np.where statement to take a pandas series as mentioned in the comments, and produce the desired output shown below:

Desired Output:

    Entries    Conflict
0    "manaaa"    "Yes"
1    "guyaaa"    "Yes"
2    "boyaaa"    "Yes"
3    "girlbbb"   "No

Answer 1

As the question was in regard to the apply function to a dataframe column for the same row, it seems more accurate to use the pandas apply funtion in combination with lambda :

import pandas as pd
df = pd.DataFrame({'number': [10, 20 , 30, 40], 'condition': ['A', 'B', 'A', 'B']})

def func(number,condition):
    multiplier = {'A': 2, 'B': 4}
    return number * multiplier[condition]

df['new_number'] = df.apply(lambda x: func(x['number'], x['condition']), axis=1)

In this example, lambda takes the columns 'number' and 'condition' of the dataframe df and applies these columns of the same row to the function func with apply .

This returns the following result:

df
Out[10]: 
 condition  number  new_number
0   A   10  20
1   B   20  80
2   A   30  60
3   B   40  160

For the UPDATE case its also possible to use the pandas apply function:

df1 = pd.DataFrame({'Entries':['man','guy','boy','girl'],'Conflict':['Yes','Yes','Yes','No']})

def funcA(d):
    d = d + 'aaa'
    return d
def funcB(d):
    d = d + 'bbb'
    return d

df1['Entries'] = df1.apply(lambda x: funcA(x['Entries']) if x['Conflict'] == 'Yes' else funcB(x['Entries']), axis=1)

In this example, lambda takes the columns 'Entries' and 'Conflict' of the dataframe df and applies these columns either to funcA or funcB of the same row with apply . The condition if funcA or funcB will be applied is done with an if-else clause in lambda.

This returns the following result:

df
Out[12]:


    Conflict    Entries
0   Yes     manaaa
1   Yes     guyaaa
2   Yes     boyaaa
3   No  girlbbb

Answer 2

I don't know about using pandas.DataFrame.apply , but you could define a certain condition:multiplier key-value mapping (seen in multiplier below), and pass that into your function. Then you can use a list comprehension to calculate the new number output based on those conditions:

import pandas as pd
df = pd.DataFrame({'number': [10, 20 , 30, 40], 'condition': ['A', 'B', 'A', 'B']})

multiplier = {'A': 2, 'B': 4}

def func(num, condition, multiplier):
    return num * multiplier[condition]

df['new_number'] = [func(df.loc[idx, 'number'], df.loc[idx, 'condition'], 
                     multiplier) for idx in range(len(df))]

Here's the result:

df
Out[24]: 
  condition  number  new_number
0         A      10          30
1         B      20          80
2         A      30          90
3         B      40         160

There is likely a vectorized, pure-pandas solution that's more "ideal." But this works, too, in a pinch.