Compare elements within a list-CodePudding

List = [('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

I have this list of tuples and I want to make a nested loop to compare the first elements of the tuples together, to see if they are the same. so I want to compare j100 with k150 to see if they are the same etc. for all the elements.

Then I want to create a line of output if they are the same so d120=d120 I want the output to be "Davies and Davis have similar sounding names." and Davies and davis have to be in alphabetical order.

how would I construct something like this?

CodePudding user response：

you can use a list comprehension to do this concisely:

lst = [('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

[print(f'{i[1]} and {l[1]} have similar sounding names') for i in lst for j in lst if i[1] != j[1] and i[0] == j[0]]

Output:

Davis and Davies have similar sounding names
Davies and Davis have similar sounding names

This does it for every combination. You could break this down to a more readable loop and account for the doubles if you do not want them like so:

lst = [('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

seen = []
for i in lst:
    for j in lst:
        if i[1] != j[1] and i[0] == j[0]:
            setij = {i[1], j[1]}
            if setij not in seen:
                seen.append(setij)
                print(f'{i[1]} and {j[1]} have similar sounding names')

Output:

Davis and Davies have similar sounding names

CodePudding user response：

You can use a dict to group together names with the same tag:

List = [('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

sim = {}
for item in List:
    key,name = item
    if key in sim:
        sim[key].append(name)
    else:
        sim[key]=[name]


for _,names in sim.items():
    if len(names) > 1:
        names.sort()
        print(f'{" and ".join(names)} have similar sounding names.')

Output

Davies and Davis have similar sounding names.

CodePudding user response：

Slightly modified input:

List = [('d120', 'Davi'), ('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

You could use groupby:

from itertools import groupby, combinations
from operator import itemgetter

for _, group in groupby(sorted(List), key=itemgetter(0)):
    group = [name for _, name in group]
    if len(group) > 1:
        print(" and ".join(group)   " have similar sounding names.")

Output:

Davi and Davies and Davis have similar sounding names.

If you want pairwise ouput then you could try this with combinations:

for _, group in groupby(sorted(List), key=itemgetter(0)):
    group = [name for _, name in group]
    if len(group) > 1:
        for name1, name2 in combinations(group, 2):
            print(f"{name1} and {name2} have similar sounding names.")

Output:

Davi and Davies have similar sounding names.
Davi and Davis have similar sounding names.
Davies and Davis have similar sounding names.

CodePudding user response：

You can try something like this

List = [('j100', 'Jacob'), ('k150', 'Kevin'), ('d120', 'Davis'), ('d120', 'Davies'), ('c642', 'Charles'), ('s315', 'Stephen')]

for a in range(len(List)-1):
    a_item = List[a]
    for b in range(a 1,len(List)):
        b_item = List[b]
        
        if a_item[0] == b_item[0]:
            result = [a_item[1],b_item[1]]
            result.sort()
            print(f'{result[0]} and {result[1]} have similar sounding names.')