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

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]}

CodePudding user response：

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]}

CodePudding user response：

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)

CodePudding user response：

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))

CodePudding user response：

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 ) }