Apply a function to a dataframe column element based on the value in another column for the same row?

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 in a number column, as shown below:

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

BUT, although I am applying this function to a column of a number, I want the function to also refer to the column condition, i.e. in the pseudocode:

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 include the same function in my statement applywritten above? those. referring to the value in the condition column, acting on the value in the number column?

Note. I have read the docs on np.where(), pandas.loc()and pandas.index(), but I just can't figure out how to put this into practice.

, number condition.

, :

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

: . . :

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)}

np.where, pandas, , , :

:

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

Answer 1

pandas.DataFrame.apply, condition:multiplier (. multiplier) . number :

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))]

:

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

, , pandas , "". .

Answer 2

, dataframe , pandas apply 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)

lambda '' df func apply.

:

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

UPDATE case pandas apply:

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)

lambda "" "" df funcA, funcB apply. , funcA funcB , if-else lambda.

:

df
Out[12]:


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