Pandas .apply with conditional if in different columns

Question

I have a dataframe as below. I am trying to check if there is a nan in the Liq_Factor, if yes, put 1 otherwise divide use/TW. Result in column Testing.

+---+------------+------------+--------+--------+--------+
| 1 |            | Liq_Factor | Zscire | Use    | Tw     |
| 2 | 01/10/2020 | 36.5       | 44     | 43.875 | 11.625 |
| 3 | 02/10/2020 | Nan        | 43.625 | 13.625 | 33.25  |
| 4 | 03/10/2020 | 6.125      | 47.875 | 22.5   | 4.625  |
| 5 | 04/10/2020 | Nan        | 34.25  | 37.125 | 36     |
| 6 | 05/10/2020 | 43.875     | 17.375 | 5.5    | 36.25  |
| 7 | 06/10/2020 | 40         | 14.125 | 21.125 | 14.875 |
| 8 | 07/10/2020 | 42.25      | 44.75  | 21.25  | 31.75  |
+---+------------+------------+--------+--------+--------+

I was wondering if i can use .apply in the sense of

DF1['Testing']=(DF1['Liq_Factor'].apply(lambda x: x=1 if pd.isna(DF1['Zscore']) else DF1['Use']/DF1['Tw'])

Can you please help?

Thanks, H

gtomer · Accepted Answer · 2020-09-07 13:36:28Z

2

You can use apply or another alternative is where function from numpy:

df['Liq_Factor'] = np.where(df['Liq_Factor'] == np.Nan, 1, df['Use']/df['TW'])

Following your comments below you can do:

# create another column with the calculation
df['calc'] = (1/3)* df['ATV']/df['TW']*100000000
# create two rules (you can use one rule and then the opposite)
mask_0 = (df['calc'] < 1)
mask_1 = (df['calc'] > 1)
# change result value by condition
df.loc[mask_0, 'Liq Factor'] = df['calc']
df.loc[mask_1, 'Liq Factor'] = 1

edited Sep 7, 2020 at 13:36

answered Sep 5, 2020 at 17:37

gtomer

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10 Comments

Hassanov Over a year ago

Thank you so much! It works but i have two questions please. 1. Is there a way to do it with pandas? 2. If yes, What if i have more than one condition? Can this be done in pandas?

Sushant Over a year ago

For your first question posted the answer using pandas, for second question you can create a function with your input columns and calculation logic with many conditions, call that function using apply.

Hassanov Over a year ago

When I added the "min" I get the below error ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all(). My line code is as below Final['Liq Factor'] = np.where(Final['ATV'] == np.nan, 1, min((1/3)*(Final['ATV']/Final['TW']*1000000000),1)) Any help please?

gtomer Over a year ago

You can not do it like this, because df['ATV'] for example, is not a single number but a whole column. Let me write a solution for you. A few minutes please

Sushant Over a year ago

@Hassanov use min(a,b) to get the minimum, its working fine for me- I have updated it in my answer as well.

|

Sushant · Accepted Answer · 2020-09-07 16:08:07Z

1

Use below code-

df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan'  else x['Use']/x['Tw'], axis=1)

Based on changes in comment section

df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan'  else min(x['Use']/x['Tw'],1), axis=1)

edited Sep 7, 2020 at 16:08

answered Sep 5, 2020 at 18:16

Sushant

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