Python：列表清单-元素的存在

Question

I am trying to check the presence of an element in a list of lists and, if so, do something at this particular list (within the list of list):

transac1 = ['John','6', '20/10/2016']
transac2 = ['Emma','6', '20/10/2016']
transactions = [['Marie',2],['Emma',9]]

I would like to do the following:

## non-Python code
if ['John',x] exists in transactions:
     ## I need to have the index where [John,x] is at that point
     then transactions[index][1] += transac1[1] 
else:
     transactions.append(['John',6])

So executing this loop with transac1 would make:

transactions = [['Marie',2],['Emma',9],['John',6]]

And executing this loop with transac2 would make:

transactions = [['Marie',2],['Emma',15],['John',6]]

The problem I face with a "classic double loop" is that, each time it does not find ['John',x] it will append to the list, where I need to know that for the entire list before doing something (plus, I have the assurance that if 'John' is in the list, it is only once).

My constraint is I cannot use Dictionaries. Thanks a lot.

Answer 1

To loop over the list of list and get the indexes:

for idx, item in enumerate(transactions):
    if item[0] == 'John' and item[1] == x:
        pass
    else:
        transactions.append(['John',6])

Answer 2

Is using numpy an option? If so, you can do the following:

import numpy as np

transac1 = ['John','6', '20/10/2016']
transac2 = ['Emma','6', '20/10/2016']
transactions = [['Marie',2],['Emma',9]]

t1 = np.array(transac1)
t2 = np.array(transac2)
tt = np.array(transactions)

names = tt[:, 0]
amounts = tt[:, 1]

if t1[0] in names:
    tt[names.index(t1[0]), 1] += t[1]
else:
    tt.append([t1[0], t1[1])

If not, I would just check the name every iteration.

transac1 = ['John','6', '20/10/2016']
transac2 = ['Emma','6', '20/10/2016']
transactions = [['Marie',2],['Emma',9]]

# Run with transac1
def func():
    for i, t in enumerate(transactions):
        if t[0] == transac1[0]:
            transactions[i][1] += transac1[1]
            return transactions
    transactions.append([transac1[0], transac1[1])
    return transactions

func()