Question

Creating a Dictionary (python 3)

Description

Larry is an immigrant in the USA and has a hard time understanding English there. So he decides to make a software that will tell him the synonyms of the word that he types. He has asked you for help.

Remember, you will first need to choose a data structure that you will use to store the information about the words. You can use lists or dict or tuple or anything else for this purpose.

There will be two actions involved:

Action 1 = Store the following words as synonyms of each other.

Action 2 = Always followed by one word. Print ALL the synonyms of that word in alphabetical order (note: the synonyms printed should not have the query word).

Input:

The input will be one list whose first entry is an int ‘test’ and all other subsequent entries will be lists. ‘test’ will represent the number of lists that follow.

Each of the following lists will consist of a number as its zeroth index (1 or 2) denoting the action to be taken on the words.

Output:

Each action 2 will print the synonyms in the order it is called.

Sample input:

[ 7,

[1,'good','sound','first-class'],

[1,'enough', 'abundant'],

[1,'adequate', 'enough', 'ample'],

[2, 'adequate'],

[2, 'good'],

[1, 'bright', 'illuminous'],

[2, 'cs'] ]

Sample output:

adequate: abundant,ample,enough
good: first-class,sound
cs: 

Explanation:

There are three queries for Action 2 and, hence, the output will have three lines.

Line 1: Adequate has three synonyms ( [1,'enough', 'abundant'], [1,'adequate', 'enough', 'ample'],).

So the three synonyms will be printed in alphabetical order:

Enough: abundant,ample,enough

Line 2: Good has two synonyms ( [1,'good','sound','first-class'] ).

So the two synonyms will be printed in alphabetical order:

Good: first-class, sound

Line 3: there are no synonyms added for cs and hence no synonyms will be printed, but just “cs: ”.(Note space after cs: )
Note that the print order is the same as the order in which ACTION2 is called i.e. adequate, good, cs

Answer 1

#Test case

test = [ 7,

[1,'good','sound','first-class'],

[1,'enough', 'abundant'],

[1,'adequate', 'enough', 'ample'],

[2, 'adequate'],

[2, 'good'],

[1, 'bright', 'illuminous'],

[2, 'cs'] ]

#Function for calculating the Values

#input: values of a key,Synonym dictionary,key

def common(test_values,Syn,t):

#list for having values of keys and Here some values to the corresponding key

#also has values

#such values are stored in a

a = []

for val in test_values:

#taking values of the values of given key

a.append(list(set(Syn[val])^set(test_values)))

#taking union of all the values in the above list

b = (list(set().union(*a)))

#Remove the key as it repeats

b.remove(t)

#sort the array as it is given in question to alphabetically sort

b.sort()

#return the values

return(b)

#Function for creating the Synonyms dictionary

def makelist(test_key,values):

#Here Some keys may already exists

#So first check whether the key is present

try:

#if key exists take its values

l = Synonyms[test_key]

except KeyError:

#if that key does not exist make a new list for the values

l = []

#Append the corresponding values into the list

for x in values:

if x != 1 and x != test_key:

l.append(x)

return l

#initiialize a dictionary for Synonyms

Synonyms = dict()

#Main function to check for print or store in dicitonary

for i in range(1,test[0]+1):

if test[i][0] == 1:

#Make the dictionary

for j in range(1,len(test[i])):

Synonyms[str(test[i][j])] = makelist(test[i][j],test[i])

try:

#Print the dictionary for given key

if test[i][0] == 2:

out = common(Synonyms[str(test[i][1])],Synonyms,test[i][1])

#Here join(map) is used to print list as a values shown in output

print(str(test[i][1]) +' : ' + ','.join(map(str,out)))

#If the given key does'nt exists print a space ' '

except KeyError:

print(test[i][1] + ' : ')

#output: