Pass Dataframe Row Values Iteratively (loop) as Arguments in a Function in Python

Question

I have a function that requires three arguments:

def R0(confirm, suspect,t):
    p = 0.695
    si = 7.5
    yt = suspect * p + confirm
    lamda = math.log(yt)/t
    R0 = 1 + lamda * si + p * (1 - p) * pow(lamda * si,2)   
    return R0

And a dataframe with three columns:

data = {'confirm':  ['41', '41', '43', '44'],
        'suspect': ['0', '0', '0', '10'],
        't': ['0', '1', '2', '3']
        }

df = pd.DataFrame (data, columns = ['confirm','suspect', 't'])

I would like to use each row (with three columns, and hence three values) as the argument values for the function. Finally, I would like to loop over rows of the dataframe and return a list.

For instance, the results should look like:

result = [R0_Value1, R0_Value2, R0_Value3, ....] where

R0_Value1 = R0(41, 0, 0)
R0_Value2 = R0(41, 0, 1)
R0_Value3 = R0(43, 0, 2)
...

I figure out it probably has something to do with pandas.DataFrame.apply and *. But I am new to Python and could not figure out how to do it. Could someone please help?

Should it really be in the dataframe though? Much easier and I think a bit faster if you process outside the dataframe. — WGS
– WGS, Commented Feb 8, 2020 at 9:20
Hi Jerome, I am not sure how to process outside the dataframe. The dataframe is actually loaded from a csv file. A friend gives me a hint and I posted an answer that solves the answer. Thank you for your fast response. I really appreciate it. — June
– June, Commented Feb 8, 2020 at 9:28

Araldo van de Kraats · Accepted Answer · 2020-02-08 09:47:28Z

3

You were looking in the right direction with 'apply':

# Convert values to int (now strings, which will throw an error in R0)
df = df.applymap(int)

df['results'] = df.apply(lambda x: R0(x.confirm, x.suspect, x.t), axis=1)

What happens when you use the apply function is that (in case of axis=1) the whole row is used as the first argument in the specified function. The lambda function is basically a wrapper that transforms this single argument (x) into the three unpacked values and passes them in the correct order to the next function, R0.

answered Feb 8, 2020 at 9:47

Araldo van de Kraats

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Comments

Georgina Skibinski · Accepted Answer · 2020-02-08 10:32:56Z

You can do:

df["formula"]=df.apply(lambda x: R0(*x), axis=1)

The whole thing (there were couple of other things in need of polishing):

import pandas as pd
import math

def R0(confirm, suspect,t):
    p = 0.695
    si = 7.5
    yt = suspect * p + confirm
    lamda = math.log(yt)/max(t,1) #you need to handle division by 0 somehow
    R= 1 + lamda * si + p * (1 - p) * math.pow((lamda * si),2)
    return R

data = {'confirm':  ['41', '41', '43', '44'],
        'suspect': ['0', '0', '0', '10'],
        't': ['0', '1', '2', '3']
        }

df = pd.DataFrame(data, columns = ['confirm','suspect', 't']).astype(int) #note it has to be numeric to conduct all the arithmetics you are doing later

df["formula"]=df.apply(lambda x: R0(*x), axis=1)

Outputs:

   confirm  suspect  t     formula
0       41        0  0  193.285511
1       41        0  1  193.285511
2       43        0  2   57.274157
3       44       10  3   31.297989

Henry Yik · Accepted Answer · 2020-02-08 09:56:37Z

1

If you insist of using pandas, you can also do the calculations directly using numpy without a function:

df = pd.DataFrame (data, columns = ['confirm','suspect', 't']).astype(int)

p = 0.695
si = 7.5

df['results'] = 1 +(np.log(df["suspect"]*p + df["confirm"])/df["t"])*si \
                  + p*(1-p)*np.power((np.log(df["suspect"]*p + df["confirm"])/df["t"])*si,2)
print (df)

#
   confirm  suspect  t     results
0       41        0  0         inf
1       41        0  1  193.285511
2       43        0  2   57.274157
3       44       10  3   31.297989

answered Feb 8, 2020 at 9:56

Henry Yik

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Comments

June · Accepted Answer · 2020-02-08 09:25:44Z

0

df.apply(lambda x: R0(x[0], x[1], x[2]), axis=1) will give the right result.

answered Feb 8, 2020 at 9:25

June

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1 Comment

WGS Over a year ago

It will, only if you make sure your dataframe is using numerical values. Your values above are strings, which will throw off your apply. Additionally, it will encounter an infinite value on the first row because of division by zero.

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