Question

#importing file

users = pd.read_table('u.user', sep='|', index_col='user_id')

Describe and show the dataframe

In [ ]:

# describe information of all columns

​

# describe information of all numeric columns only

​

# describe information of all object columns only

​

# show first 10 rows of users dataframe

detecting duplicate rows

In [10]:

# check wheather a row is identical to a previous row

​

# count all duplicate rows in the dataframe

​

# show only duplicate rows in the dataframe

​

# drop all duplicate rows in the dataframe

​

# check a single specific column for duplicates occur or not

​

# check specify more than one column for finding duplicates

In [11]:

# display the 3 most frequent occupations in 'users'

​

# change the data type of a column name age from int to float

​

# for each occupation, calculate the minimum and maximum ages

​

In [12]:

# for each occupation in 'users', count the number of occurrences

​

# plot barchar of upper out w.r.t each occupation 

​

In [13]:

# for each occupation, calculate the mean age

​

# plot pie chart of the upper output

​

In [14]:

# for each combination of occupation and gender, calculate the mean age

​

# plot barchar of upper out w.r.t each occupation and gender 

​

In [15]:

# sort 'users' by 'occupation' and then by 'age' (in a single command)

u.user data set

user_id|age|gender|occupation|zip_code
1|24|M|technician|85711
2|53|F|other|94043
3|23|M|writer|32067
4|24|M|technician|43537
5|33|F|other|15213
6|42|M|executive|98101
7|57|M|administrator|91344
8|36|M|administrator|05201
9|29|M|student|01002
10|53|M|lawyer|90703
11|39|F|other|30329
12|28|F|other|06405
13|47|M|educator|29206
14|45|M|scientist|55106
15|49|F|educator|97301
16|21|M|entertainment|10309
17|30|M|programmer|06355
18|35|F|other|37212
19|40|M|librarian|02138
20|42|F|homemaker|95660
21|26|M|writer|30068
22|25|M|writer|40206
23|30|F|artist|48197

Answer 1

#importing file
import pandas as pd
import numpy as np
users = pd.read_table('user.csv', sep='|', index_col='user_id')

# describe information of all columns
print(users.describe())

# describe information of all numeric columns only
print(users.describe(include=[np.number]))

# describe information of all object columns only
print(users.describe(include=[object]))

# show first 10 rows of users dataframe detecting duplicate rows
print(users.head(10))

# check wheather a row is identical to a previous row

# count all duplicate rows in the dataframe
print(users.pivot_table(index = ['age', 'gender','occupation','zip_code'], aggfunc ='size') )

# show only duplicate rows in the dataframe

# drop all duplicate rows in the dataframe

# check a single specific column for duplicates occur or not
print(users.pivot_table(index = [ 'gender'], aggfunc ='size') )

# check specify more than one column for finding duplicates
print(users.pivot_table(index = ['occupation','gender'], aggfunc ='size') )

# display the 3 most frequent occupations in 'users'
print(users['occupation'].value_counts()[:3])

# change the data type of a column name age from int to float
convert_col = {'age': float}
users = users.astype(convert_col)
print(users.dtypes)

# for each occupation, calculate the minimum and maximum ages
print(users.groupby(['occupation'])['age'].agg({'Min N Max':['min', 'max']}))

# for each occupation in 'users', count the number of occurrences
output = users['occupation'].value_counts()[:]
print(output)

# plot barchar of upper out w.r.t each occupation
output.plot.bar(rot=0)

# for each occupation, calculate the mean age
output=users.groupby(['occupation'])['age'].agg({'':['mean']})
print(output)

# plot pie chart of the upper output
output.plot.pie(subplots=True, figsize=(10,10))

# for each combination of occupation and gender, calculate the mean age
output=users.groupby(['occupation','gender'])['age'].agg({'':['mean']})
print(output)

# plot barchar of upper out w.r.t each occupation and gender
output.plot.bar(rot=0)

# sort 'users' by 'occupation' and then by 'age' (in a single command)
print(users.sort_values(by=['occupation', 'age']))