CODEWITHSUNDEEP
python
Maria Nazari
Maria Nazari

Reputation: 690

IF Statement Dependent on Loop

I have the following script:

import pandas as pd

ls = [
      ['A', 1, 'A1', 9],
      ['A', 1, 'A1', 6],
      ['A', 1, 'A1', 3],
      ['A', 2, 'A2', 7],
      ['A', 3, 'A3', 9],
      ['B', 1, 'B1', 7],
      ['B', 1, 'B1', 3],
      ['B', 2, 'B2', 7],
      ['B', 2, 'B2', 8],
      ['C', 1, 'C1', 9],

      ]

#convert to dataframe
df = pd.DataFrame(ls, columns = ["Main_Group", "Sub_Group", "Concat_GRP_Name", "X_1"]) 

#get count and sum of concatenated groups
df_sum = df.groupby('Concat_GRP_Name')['X_1'].agg(['sum','count']).reset_index()

#print in permutations formula to calculate different permutation combos   
import itertools as it
perms = it.permutations(df_sum.Concat_GRP_Name)


def combute_combinations(df, colname):
    l = []
    import itertools as it
    perms = it.permutations(df[colname])

    for perm_pairs in perms:
        #take in only the first three pairs of permuations and make sure
        #the first column starts with A, secon with B, and third with C
        if 'A' in perm_pairs[0] and 'B' in perm_pairs[1] and 'C' in perm_pairs[2]:
            l.append([perm_pairs[0], perm_pairs[1], perm_pairs[2]])
    return l

#apply function, this will generate a list of all of the permuation pairs
t = combute_combinations(df_sum, 'Concat_GRP_Name' )

#convert to dataframe and drop duplicate pairs
df2 = pd.DataFrame(t, columns = ["Item1", 'Item2', 'Item3']) .drop_duplicates()

I am not sure how to combine the components of a loop inside of an IF statement. From the example above, I knew that I had three different types of the Main_Group variable. Let's say that I didn't know how many unique values existed in the Main_Group column. How do I update the following IF statement to account for this?

if 'A' in perm_pairs[0] and 'B' in perm_pairs[1] and 'C' in perm_pairs[2]:
                l.append([perm_pairs[0], perm_pairs[1], perm_pairs[2]])

I want each of the variable in it's own column. If I have 5 types of main group then I will have perm_pairs[0] to perm_pairs[4] in my IF statement. I was thinking about extracting the values in the Main_Group and turning this into a set. Then I would iterate through each value and use it's length to figure out the IF statement, but so far the logic is not working out. How do I iterate through the set and then update the IF statement?

Upvotes: 0

Views: 85

Answers (1)

Richard Nemeth
Richard Nemeth

Reputation: 1864

To make the condition more dynamic, you can refactor your function like this:

import numpy as np

def combute_combinations(df, colname, main_group_series):
    l = []
    import itertools as it
    perms = it.permutations(df[colname])

    # Provides sorted list of unique values in the Series
    unique_groups = np.unique(main_group_series)

    for perm_pairs in perms:
        #take in only the first three pairs of permuations and make sure
        #the first column starts with A, secon with B, and third with C
        if all([main_group in perm_pairs[ind] for ind, main_group in enumerate(unique_groups)]):
            l.append([perm_pairs[ind] for ind in range(unique_groups.shape[0])])
    return l

Then you can just call the function as before, but include the series of the main group column

t = combute_combinations(df_sum, 'Concat_GRP_Name', df['Main_Group'])

Upvotes: 1

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