CODEWITHSUNDEEP
pythonpandas
Magellan88
Magellan88

Reputation: 2573

Pandas merging a Dataframe and a series

I have two pandas DataFrames: one containing the anonymous hashes of customer numbers (here for simplicity the hashes of the numbers 0-19)

import pandas as pd
import numpy as np
from hashlib import sha1

df_customers = pd.DataFrame( [ sha1(i).hexdigest() for i in     np.arange(20)] )
df_customers.columns = ["customer"]

Now I have a second table (DataFrame) with 200 occurrences of customers picking from a selection of 20 different kinds of fruit:

fruit = ["apple", "banana", "peach", "plum", "orange", "cumquat", "raspberry", "lemon", "rubarb", "pineapple"]
pd.DataFrame( np.c_[ np.array([ sha1(i).hexdigest() for i in np.random.randint(0,20,200) ]),
          np.array([ fruit[i]  for i in np.random.randint(0,len(fruit),200) ]) ],
         columns=("customer_id","fruit")
         )

now I would like to add a column to the customer DataFrame that indicated the variety of eaten fruit -- that is the number of different fruit that have been eaten by each customer. For that I did:

variety = df_eating.groupby("customer_id")["fruit"].apply( lambda x: len(np.unique(x)))

this gives me a "Series". Now I feel there should be a straight forward way to add this back to df_customer, respecting the customer_id but here I'm quite stuck:

pd_customer["variety"] = variety

does not respect the customer id and gives NaN for every value

and functions like pd.merge() where there is an option to merge "on" something did not do what I wanted.

Upvotes: 2

Views: 629

Answers (1)

EdChum
EdChum

Reputation: 393963

If I understand what you want then you can call map and pass the series:

In [36]:

df_customers['variety'] = df_customers['customer'].map(variety)
df_customers
Out[36]:
                                    customer  variety
0   9069ca78e7450a285173431b3e52c5c25299e473        7
1   3c585604e87f855973731fea83e21fab9392d2fc        9
2   0aaf76f425c6e0f43a36197de768e67d9e035abb        6
3   8e146c3c4e33449f95a49679795f74f7ae19ecc1        6
4   d6459ab29c7b9a9fbf0c7c15fa35faa30fbf8cc6        7
5   ddaf0ed54dfc227ce677b5c2b44e3edee7c7db77        5
6   8098e7dfb09adba3bf783794ba0db81985a814d7        6
7   2f086fc767a0dac59a38c67f409b4f74a1eab39f        8
8   a454ca483b4a66b83826d061be2859dd79ff0d6c        7
9   9db063f3b5e0adfd0d29a03db0a1c207b3740a94        6
10  eb408ddc4fa484e6befdf5954e56a2198c7a9fab        8
11  94312fc592ee3f323b3f9d8612737c507ec7f6c3        5
12  f3a56292ca640b843071c9a143404cea014f4d5c        9
13  b1197c208248d0f7ffb3e322d5ec187441dc1b26        7
14  f143c36fc53bfde11a8d122249aced46c43cc2e2        7
15  aefa2f5632d36978838bff3aabcef5ee01395729        5
16  5497b0911b3f5772723def3b360a2e654327c19b        6
17  498bcbf6cbffcc8dd2623f388d81f44cfad1014d        5
18  96760d655a51e69d67d32a5f18c23c9bfe0576cf        5
19  fe5aa6438ae9b661b033b91e9c679ad2898cbfd4        6

With regards to optimising your code you can replace this line:

variety = df_eating.groupby("customer_id")["fruit"].apply( lambda x: len(np.unique(x)))

with the equivalent:

variety = df_eating.groupby("customer_id")["fruit"].nunique()

Upvotes: 4

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