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Create Pandas Dataframes From Unique Values In One Column

I have a Pandas dataframe with 1000s of rows. and it has the Names column includes the customer names and their records. I want to create individual dataframes for each customer ba

Solution 1:

Your current iteration overwrites x twice every time it runs: the for loop assigns a customer name to x, and then you assign a dataframe to it.

To be able to call each dataframe later by name, try storing them in a dictionary:

df_dict = {name: df.loc[df['customer name'] == name] for name in customerNames}

df_dict['Name3']

Solution 2:

To create a dataframe for all the unique values in a column, create a dict of dataframes, as follows.

  • Creates a dict, where each key is a unique value from the column of choice and the value is a dataframe.
  • Access each dataframe as you would a standard dict (e.g. df_names['Name1'])
  • .groupby() creates a generator, which can be unpacked.
    • k is the unique values in the column and v is the data associated with each k.

With a for-loop and .groupby:

df_names = dict()
for k, v in df.groupby('customer name'):
    df_names[k] = v

With a Python Dictionary Comprehension

Using .groupby

df_names = {k: v for (k, v) in df.groupby('customer name')}
  • This comes from a conversation with rafaelc, who pointed out that using .groupby is faster than .unique.
    • With 6 unique values in the column, .groupby is faster, at 104 ms compared to 392 ms
    • With 26 unique values in the column, .groupby is faster, at 147 ms compared to 1.53 s.
  • Using an a for-loop is slightly faster than a comprehension, particularly for more unique column values or lots of rows (e.g. 10M).

Using .unique:

df_names = {name: df[df['customer name'] == name] for name in df['customer name'].unique()}

Testing

  • The following data was used for testing
import pandas as pd
import string
import random

random.seed(365)

# 6 unique values
data = {'class': [random.choice(['1-5', '6-25', '26-100', '100-500', '500-1000', '>1000']) for _ inrange(1000000)],
        'treatment': [random.choice(['Yes', 'No']) for _ inrange(1000000)]}

# 26 unique values
data = {'class': [random.choice( list(string.ascii_lowercase)) for _ inrange(1000000)],
        'treatment': [random.choice(['Yes', 'No']) for _ inrange(1000000)]}

df = pd.DataFrame(data)

Solution 3:

maybe i get you wrong but

when

forxin customerNames:
    x = DataFrame.loc[DataFrame['customer name'] == x]
x

gives you the right output for the last list entry its because your output is out of the indent of the loop

import pandas as pd

customer_df = pd.DataFrame.from_items([('A', ['Jean', 'France']), ('B', ['James', 'USA'])],
                        orient='index', columns=['customer', 'country'])

customer_list = ['James', 'Jean']

for x in customer_list:
    x = customer_df.loc[customer_df['customer'] == x]
    print(x)
    print('now I could append the data to something new')

you get the output:

  customer country
B    James     USA
now I could append the data to something new
  customer country
A     Jean  France
now I could append the data to something new

Or if you dont like loops you could go with

import pandas as pd

customer_df = pd.DataFrame.from_items([('A', ['Jean', 'France']), ('B', ['James', 'USA']),('C', ['Hans', 'Germany'])],
                        orient='index', columns=['customer', 'country'])

customer_list = ['James', 'Jean']


print(customer_df[customer_df['customer'].isin(customer_list)])

Output:

  customer country
A     Jean  France
B    James     USA

df.isin is better explained under:How to implement 'in' and 'not in' for Pandas dataframe

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