Question

Hi, I have a python coding questions and would need help:

Count the neighbors of each node in a graph. input graph is a multi-dimensional list

Let's say I want to have each person's friends count:

input (list of list): [[A, B], [B, C], [B, D], [E]],

output (dict): {A:1, B: 3, C:1, D:1, E:0}

assume you won't get a repeat pair like [A,B] and [B,A], and neither will there be more than 2 people in a relationship.

I want to write a function that creates a dictionary of how many friends each person has. People can have 0 to many friends.

Answer 1

Below is the Python code I hope that i have provided sufficient comments for your better understanding Note that I have done proper indentation but this code is automatically left alligned on this interface

def friends(lstOfList):   
d = {} #empty dictionary
for lst in lstOfList: #Traverse every list inside 2d list
if len(lst)==2: #Atleast 2 friends are present in this list
if lst[0] in d: #This person is already present in dictionary
d[lst[0]]+=1 #increment friend of 1st entry by 1
else:
d[lst[0]]=1 #This person appeared first time hence initialize it by 1

if lst[1] in d: #This person is already present in dictionary
d[lst[1]]+=1 #increment friend of 1st entry by 1
else:
d[lst[1]]=1 #This person appeared first time hence initialize it by 1
else: #single entry in this list
if lst[0] not in d: #This person is not present in dictionary
d[lst[0]]=0 #This person(single) appeared first time hence initialize it by 0
return d

print(friends([['A', 'B'], ['B', 'C'], ['B', 'D'], ['E']]))

test.py- CA Windows System321test.py (3.5.0) File Edit Format Run Options Window Help friends (1st0fLǐst): d-{} #empty dictionary def for ïst ïn ïst0fList : #Traverse every list ins ide 2d list if len(1st) #Atleast 2 friends are present in this list --2: in if #This person is already present in dictionary 1st [O] d(1st [0] ]+-1 d: #increment friend of 1st entry by 1 else: d [1st [O]-1 #This person appeared first time hence initialize it by 1 if 1st [1] in d: #This person is already present in dictionary d(1st [1]]+-1 #increment friend of 1st entry by 1 else: d(1st [1]1-1 #This person appeared first time hence initialize it by 1 else : #single entry in this list if lst [O] not in d: #This person is not present in dictionary d(1st [0]]=0 #This person (single) appeared first time hence initialize it by 0 return d print (friends (['A, 'B', ['B', '', ['B', 'D'], ['E))

Below is the screenshot of output

Python 3.5.0 Shell File Edit Shell Debug Options Window Help Python 3.5.0 (v3.5.0: 374f501f4567, Sep 13 2015, 02:16:59) [MSC v.1900 32 bit (Intel) on win32 Type "copyright", "credits" or "license"for more information.

I have tried to explain it in very simple language and I hope that i have answered your question satisfactorily.Leave doubts in comment section if any.