CODEWITHSUNDEEP
pythonlistdictionary
hongyin
hongyin

Reputation: 35

How to use to list to create a dictionary has multiple value

I am creating a dictionary using two lists that have multiple values, and without using zip.

If I have two lists like this:

store1=['mango,5', 'apple,10', 'banana,6']
store2=['mango,7', 'apple,8', 'banana,7']

How do I create a dictionary like this:

dic={'mango':[5,7],'apple':[10,8],'banana':[6,7]}

Upvotes: 2

Views: 65

Answers (6)

Ashley
Ashley

Reputation: 628

Variation #1...

from collections import defaultdict
r = defaultdict(list)
for s, i in [item.split(',') for item in store1 + store2]: r[s].append(int(i))

Variation #2...

If I were planning on handling raw info from one or more stores, I might have a function and I would go for readability rather than compactness, and use type hints:

def raw_store_info_to_dict(all_store_raw_info: list[str]) -> defaultdict[str,list[int]]:
    all_stores_dict = defaultdict(list)
    for s, i in [item.split(',') for item in chain.from_iterable(all_store_raw_info)]:
        all_stores_dict[s].append(int(i))
    return all_stores_dict

store1=['mango,5', 'apple,10', 'banana,6']
store2=['mango,7', 'apple,8', 'banana,7']
store3=['mango,17', 'apple,18', 'banana,17']
store4=['grape,3', 'cider,2', 'banana,77']
store5=['mango,27', 'apple,28', 'banana,37']
print(raw_store_info_to_dict([store1, store2, store3, store4, store5]))

# Output:
# defaultdict(<class 'list'>, {'mango': [5, 7, 17, 27], 'apple': [10, 8, 18, 28], 'banana': [6, 7, 17, 77, 37], 'grape': [3], 'cider': [2]})

Upvotes: 1

Hassan Shahzad
Hassan Shahzad

Reputation: 1

Use dict.setdefault:

d = {}
for i in store1 + store2:
  key,value = i.split(',')
  d.setdefault(key, []).append(int(value))

Upvotes: 0

Xin Cheng
Xin Cheng

Reputation: 1458

Without zip, ok.

st1 = sorted([ e.split(',') for e in store1 ])
st2 = sorted([ e.split(',') for e in store2 ])
dic = { st1[i][0]: [ st1[i][1], st2[i][1] ] for i in range(len(st1))}

Without zip, if the lists are not in good form. i.e. not in the correct order, or the keys may not exist on both lists, etc. This solution has a better compatibility.

dic = { k:v for [k,v] in [ e.split(',') for e in store1 ] }
for [k,v] in [ e.split(',') for e in store2 ]:
    if k in dic: dic[k].append(v)
    else: dic[k]=[v]

With zip One line:

dic = { kv1.split(',')[0]: [kv1.split(',')[1], kv2.split(',')[1]] for kv1,kv2 in zip(store1,store2) }

With zip, another solution with a better runtime performance if your data is big

st1 = [ e.split(',') for e in store1 ]
st2 = [ e.split(',') for e in store2 ]
dic = { kv1[0]: [ kv1[1], kv2[1] ] for kv1,kv2 in zip( st1, st2 ) }

Upvotes: 0

Rahul K P
Rahul K P

Reputation: 16091

You can do this with dict.setdefault,

In [2]: d = {}
   ...: for i in store1 + store2:
   ...:     key,value = i.split(',')
   ...:     d.setdefault(key, []).append(int(value)) 

In [3]: d
Out[3]: {'mango': [5, 7], 'apple': [10, 8], 'banana': [6, 7]}

Upvotes: 1

Wolric
Wolric

Reputation: 767

Adding another solution without using zip is to just concatenate the given lists thus creating one big list to iterate through:

d={}
for store in (store1 + store2):
    key, val = store.split(",")
    if key not in d:
        d[key] = []
    d[key].append(int(val))

Upvotes: 0

Sanjay SS
Sanjay SS

Reputation: 566

You could try using 2 separate for-loops to add them to the dictionary

    store1 = ['mango,5', 'apple,10', 'banana,6']
    store2 = ['mango,7', 'apple,8', 'banana,7']
    d = {}
    for i in store1:
        a, b = i.split(',')
        if a not in d:
            d[a] = []
        d[a].append(int(b))
    for i in store2:
        a, b = i.split(',')
        if a not in d:
            d[a] = []
        d[a].append(int(b))
    print(d)

Or Use a nested for-loop

    store1 = ['mango,5', 'apple,10', 'banana,6']
    store2 = ['mango,7', 'apple,8', 'banana,7']
    d = {}
    for i in [store1,store2]:
        for j in i:
            a, b = j.split(',')
            if a not in d:
                d[a] = []
            d[a].append(int(b))
    print(d)

Upvotes: 0

Related Questions