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pythonpandas
Lincoln Hannah
Lincoln Hannah

Reputation: 3

Python Pandas Dataframe - row level operations

I need to do a large number of row level operations (a few pages of code) on a table of data.

E.g. if row.Col_A == 'X' : row.Col_B = 'Y'

I believe iterrows isn't appropriate for altering table values. So I've converted the table to a list of DotMap dictionaries. With this I can loop over the list and for each dictionary (row), write the code as above and the alterations are saved.

Is it possible to do this with the data as a DataFrame ?

There is a lot of logic and I think its clearest written this way so I'd prefer not to use map or apply functions.

Upvotes: 0

Views: 1156

Answers (3)

cyau
cyau

Reputation: 449

Let's have the following example dataframe:

import pandas as pd
import numpy as np

some_data = pd.DataFrame({
    'col_a': [1, 2, 1, 2, 3, 4, 3, 4],
    'col_b': ['a', 'b', 'c', 'c', 'a', 'b', 'z', 'z']
})

We want to create a new column based on one (or more) of the existing columns' values.

In case you have only two options, I would suggest using numpy.where like this:

some_data['np_where_example'] = np.where(some_data.col_a < 3, 'less_than_3', 'greater_than_3')
print(some_data)
>>>
   col_a col_b           col_c map_example np_where_example  \
0      1     a     less_than_3         NaN      less_than_3   
1      2     b     less_than_3         BBB      less_than_3   
2      1     c     less_than_3         NaN      less_than_3   
3      2     c     less_than_3         NaN      less_than_3   
4      3     a  greater_than_3         NaN   greater_than_3   
5      4     b  greater_than_3         BBB   greater_than_3   
6      3     z  greater_than_3         ZZZ   greater_than_3   
7      4     z  greater_than_3         ZZZ   greater_than_3 

# multiple conditions
some_data['np_where_multiple_conditions'] = np.where(((some_data.col_a >= 3) & (some_data.col_b == 'z')),
                                                     'is_true',
                                                     'is_false')
print(some_data)
>>>
   col_a col_b np_where_multiple_conditions
0      1     a                     is_false
1      2     b                     is_false
2      1     c                     is_false
3      2     c                     is_false
4      3     a                     is_false
5      4     b                     is_false
6      3     z                      is_true
7      4     z                      is_true

In case you have many options, then pandas.map would be better:

some_data['map_example'] = some_data.col_b.map({
    'b': 'BBB',
    'z': 'ZZZ'
})
print(some_data)
>>>
   col_a col_b map_example
0      1     a         NaN
1      2     b         BBB
2      1     c         NaN
3      2     c         NaN
4      3     a         NaN
5      4     b         BBB
6      3     z         ZZZ
7      4     z         ZZZ

As you see, in all cases the values for which a condition is not specified evaluate to NaN.

Upvotes: 1

Serge Ballesta
Serge Ballesta

Reputation: 148880

A possible way to iterate over a dataframe by rows and change column values is:

  1. make sure that there are no duplicated values in index (if there are, just use reset_index to get an acceptable index)

  2. iterate over the index and access the individual values with at

     for ix in df.index:
     if df.at[ix, 'A'] == ...:
         df.at[ix, 'B'] = z

Alternatively, if you can access the columns by their positions instead of their names, you can use the even more efficient iat:

for i in range(len(df)):
    if df.iat[i, index_col_A] == ... :
        df.iat[i, index_col_B] = z

As you access directly the individual elements, you avoid the overhead of iterrows creating a Series per row, and can perform changes. AFAIK, it is the less bad way when you cannot use the vectorized Pandas or numpy methods.

Upvotes: 0

Erik
Erik

Reputation: 867

You can use the apply function with a lambda in the following way:

df['Col_B'] = df['Col_A'].apply(lambda a: 'Y' if a == 'X' else 'N')

This creates the column Col_B on the dataframe df by looking at Col_A and giving either the values 'Y' if Col_A is 'X' and 'N' otherwise.

if your function is a bit more complex you can define it beforehand and call it in the apply function as follows:

def yes_or_no(x):
    if x == 'X':
        return 'Y'
    else:
        return 'N'
df['Col_B'] = df['Col_A'].apply(lambda a: yes_or_no(a))

Upvotes: 0

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