I have list:
['Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII',
'Security',
'Airbag']
how do I turn it into dictionary? I'm using different type of list than what is described in problem here. Also some list elements are empty, I need to omit them in final dictionary.
If your pattern always like you depict, you can split(': ') your items
my_list = ['Location: NY', 'price: 11103 USD', 'Year: 2018', 'Model: Golf VII']
my_dict = { i.split(': ')[0] : i.split(': ')[1] for i in my_list }
ADDITION FOR THE EMPTY VALUES:
You can also extract your logic into a function and make some other controls (trimming, null values etc), and still use dictionary comprehension as follows:
def str_to_keyval(inp, separator=':'):
# to list
li = inp.split(separator) if (separator in inp) else [inp, '']
# strip elements
return [i.strip() for i in li]
my_list = [
'Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII',
'Security',
'Airbag'
]
my_dict = { str_to_keyval(i)[0] : str_to_keyval(i)[1] for i in my_list }
You can change the presentation of null values (ignore their keys, or denote with an empty string etc.) through editing the first line of str_to_keyval function in the example. As it is, it gives empty string as missing values.
You can use do a dict comperhension for this application,
My solution is similar to that of @vahdet's answer, but this solution will take care of multiple spaces that may occur in key or value.
l = ['Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII']
dict( (x.split(':')[0].strip(), x.split(':')[1].strip()) for x in l )
You can make use of this same dict comprehension to handle invalid items in the list by checking whether : exists in the list or not like this and adding the key-value only if the item is valid,
l = ['Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII',
'Security',
'Airbag']
print(dict((x.split(':')[0].strip(), x.split(':')[1].strip()) for x in l if ':' in x))
Ok, so the first thing you'll need to do is loop over every item in the list. This is pretty easy using a for loop
for item in lst:
# do stuff with the item
After that, you'll need to separate the item into a key and a value. The keys and values are already separated by a colon and a space, so I would just split the string using that.
for item in lst:
split_item = item.split(": ")
key = split_item[0]
value = split_item[1]
# then you need to add the key and item to the dictionary
Adding the value is pretty simple. Just define the new value:
dictionary = {} # just to initialize the dictionary, if you already initialized it, don't do this
for item in lst:
split_item = item.split(": ")
key = split_item[0]
value = split_item[1]
dictionary[key] = value
That's it
You can use the split function to get the parts of each string to create your dictionary reference:
arr = ['Location: NY', 'price: 11103 USD', 'Year: 2018', 'Model: Golf VII']
dict = {}
for i in range(0, len(arr)):
key, value = arr[i].split(": ")
dict[key] = value
print(dict)
With this list:
mylist = ['Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII']
Try following code:
mydict = {}
for l in mylist:
a,b = l.split(': ')
mydict[a] = b
print(mydict)
Output:
{'Year': '2018', 'Model': 'Golf VII', 'price': '11103 USD', 'Location': 'NY'}
You can use a simple comprehension with dict:
my_list = ['Location: NY', 'price: 11103 USD', 'Year: 2018', 'Model: Golf VII']
result = dict(i.split(': ') for i in my_list)
Output:
{'Location': 'NY', 'price': '11103 USD', 'Year': '2018', 'Model': 'Golf VII'}
a=['Location: NY',
'price: 11103 USD',
'Year: 2018',
'Model: Golf VII']
b=[]
x={}
for i in range(len(a)):
b.append(a[i].split(':'))
for i in range(len(b)):
for j in range(len(b[i])):
x[b[i][0]]=b[i][1]
print(x)
