Count the list items times of occurances

Question

Let's assume I have the following list names = ['tan','2','alp','3','tan','4','alp','3','tan','1']

the odd indexed elements are some values and even indexed elements are their number of occurrences (so alp happened 3+3 = 6 times for example)

I am trying to make code to do that.

enter code here

names = ['tan','2','alp','3','tan','4','alp','3','tan','1']

i = 1;
dd = names[0::2]
nn = names[1::2]
ct = [0 for cc in range(len(dd))];
le = len(dd);

for i in range(0, le):
    ct[i] = int(nn[i])
    for j in range(i+1, le):
        if (j < le) and (dd[i] == dd[j]):
            ct[i] += int(nn[j]);
            del(dd[j])
            # del(nn[j])
            le -= 1

the output i get for ct is [9, 7, 4, 3, 1] however it should be [7, 6] --- 7 for tan and 6 for alp

if i uncommented del(nn[j]) --- i will be equal to le and code will stop

but i should delete the element and the number of occurrences it happened any time after first time (after adding it in ct for sure) so the counting process doesn't get repeated

any idea how to do that?

Answer 1

We can use a dictionary to keep track of the counts more easily than a list. Below I use a defaultdict which is a dict subclass from the standard library that supports default values.

from collections import defaultdict

names = ['tan','2','alp','3','tan','4','alp','3','tan','1']

d=defaultdict(int)
for name, count in zip(*[iter(names)]*2):
    d[name] += int(count)

Answer 2

That's relatively easy - build a lookup map from your odd elements and then sum the matching even values. You can use collections.defaultdict() to make your life easier:

import collections

names = ['tan', '2', 'alp', '3', 'tan', '4', 'alp', '3', 'tan', '1']

lookup_map = collections.defaultdict(int)
for element, value in zip(names[0::2], names[1::2]):
    lookup_map[element] += int(value)

print(dict(lookup_map))  # {'tan': 7, 'alp': 6}

If you really need only the values and need to keep the order, you can add an additional step:

result = [lookup_map.pop(element) for element in names[0::2] if element in lookup_map]

print(result)  # [7, 6]

If you're after the occurrences / total occurrences per element calculation, you can just apply the sum-total over the whole lookup map:

sum_total = sum(lookup_map.values())
for name in lookup_map:
    lookup_map[name] /= sum_total

print(dict(lookup_map))  # {'alp': 0.46153846153846156, 'tan': 0.5384615384615384}

Answer 3

You could use un object to store the occurences of each name

names = ['tan','2','alp','3','tan','4','alp','3','tan','1']

# Use an object to store occurences
results = {}

# Iterate through 'names' to get occurences
for i in range(len(names)):
    if i % 2 == 0:
        # Store them in the object
        if names[i] in results: # if 'name' exist in results object
            results[names[i]] += int(names[i+1])
        else:                   # if it doesn't
            results[names[i]] = int(names[i+1])

# Get the value only array
res_arr = [v for k,v in results.items()]
print results
print res_arr

This would output :

{'tan': 7, 'alp': 6}
[7, 6]