dictionary comprehension python from list

stackoverflow.com › questions › 1747817 › create-a-dictionary-with-comprehension

Use a dict comprehension (Python 2.7 and later):

Use with AI{key: value for key, value in zip(keys, values)}

Alternatively, use the dict constructor:

Use with AIpairs = [('a', 1), ('b', 2)]
dict(pairs)                          # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs)  # → {'a': 11, 'b': 12}

Given separate lists of keys and values, use the dict constructor with zip:

Use with AIkeys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values))              # → {'a': 1, 'b': 2}

Answer from fortran on Stack Overflow

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stackoverflow.com › questions › 1747817 › create-a-dictionary-with-comprehension

python - Create a dictionary with comprehension - Stack Overflow

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2369

Use a dict comprehension (Python 2.7 and later):

Use with AI{key: value for key, value in zip(keys, values)}

Alternatively, use the dict constructor:

Use with AIpairs = [('a', 1), ('b', 2)]
dict(pairs)                          # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs)  # → {'a': 11, 'b': 12}

Given separate lists of keys and values, use the dict constructor with zip:

Use with AIkeys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values))              # → {'a': 1, 'b': 2}

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306

In Python 3 and Python 2.7+, dictionary comprehensions look like the below:

Use with AId = {k:v for k, v in iterable}

For Python 2.6 or earlier, see fortran's answer.

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December 4, 2024 - You can create a dictionary from a list using dictionary comprehension in Python by iterating over the list and setting the list elements as keys and values in the new dictionary.

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Python list comprehension for dictionaries in dictionaries? - Stack Overflow

That way I can simply edit my dictionary with var['test1420'] = ... But list comprehensions don't work on that! And I can't edit lists this way because you can't assign an index like that. ... List comprehensions can be cool, but don't go overboard. There's no steadfast rule against giant comprehensions, but for what it's worth, Google's Python ... More on stackoverflow.com

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How to create a dictionary with "list comprehension"?

You can't. Dictionaries can't have duplicate keys. That's one of the fundamental properties of dictionaries/maps. You're trying to have duplicate 0 keys, which isn't allowed. The opposite is allowed though: >>> { i:0 for i in range(3,6) } {3: 0, 4: 0, 5: 0} Also note, this is called a dictionary comprehension. More on reddit.com

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January 22, 2022

Create dictionary from list of dictionaries using list comprehension

You can't do it with a list comprehension because the result of a list comprehension is a list. You can do it with a dictionary comprehension, though: result = {key: value for dictionary in list_of_dictionaries for key, value in dictionary.items()} More on reddit.com

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Can you append to a dictionary with a list comprehension in Python?

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You can do this:

Use with AIs = dict([ (k,r) for k,r in mydict.iteritems() if r['x'] > 92 and r['x'] < 95 and r['y'] > 70 and r['y'] < 75 ])

This takes a dict as you specified and returns a 'filtered' dict.

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If dct is

Use with AI{'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'},
 'test277': {'y': 72, 'x': 94, 'fname': 'test277'},}

Perhaps you are looking for something like:

Use with AI[ subdct for key,subdct in dct.iteritems() 
  if 92<subdct['x']<95 and 70<subdct['y']<75 ]

A little nicety is that Python allows you to chain inequalities:

Use with AI92<dct[key]['x']<95

instead of

Use with AIif r['x'] > 92 and r['x'] < 95

Note also that above I've written a list comprehension, so you get back a list (in this case, of dicts).

In Python3 there are such things as dict comprehensions as well:

Use with AI{ n: n*n for n in range(5) } # dict comprehension
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

In Python2 the equivalent would be

Use with AIdict( (n,n*n) for n in range(5) )

I'm not sure if you are looking for a list of dicts or a dict of dicts, but if you understand the examples above, it is easy to modify my answer to get what you want.

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August 24, 2021 - We can now use Python's zip() function to zip these two lists to generate the key-value pairs. Note: The zip function takes in a sequence of iterables as the argument, and returns an iterator of tuples, as shown in the image below. So, the first tuple is the first key-value pair, the second tuple is the second key-value pair, and in general, the i-th tuple is the i-th key-value pair. In this case, the dictionary comprehension takes the following form:

reddit.com › r/learnpython › how to create a dictionary with "list comprehension"?

r/learnpython on Reddit: How to create a dictionary with "list comprehension"?

January 22, 2022 -

I'm trying to create the following dictionary:

{ 0:3,0:4,0:5 }

And I did:

{ 0:i for i in range(3,6) }

Which doesn't work. Any idea how to? Maybe I can make it through a list of pairs? But is this way possible?

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You can't. Dictionaries can't have duplicate keys. That's one of the fundamental properties of dictionaries/maps. You're trying to have duplicate 0 keys, which isn't allowed. The opposite is allowed though: >>> { i:0 for i in range(3,6) } {3: 0, 4: 0, 5: 0} Also note, this is called a dictionary comprehension.

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One option that you might be thinking of is a dictionary from int to some container, like a list or a set: { 0: [3, 4, 5] } { 0: {3, 4, 5} }

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reddit.com › r/learnpython › create dictionary from list of dictionaries using list comprehension

r/learnpython on Reddit: Create dictionary from list of dictionaries using list comprehension

November 3, 2022 -

Original data:

list_of_dictionaries = [{'key1':'value1'},{'key2':'value2'},{'key3':['item1','item2','item3']}]

Desired output

{'key1':'value1','key2':'value2','key3': ['item1','item2','item3']}

Standart loop does the job, but i don't like it

empty_dictionary = {}
for item in list_of_dictionaries:
    for k, v in item.items():
        empty_dictionary[k] = v

How can i do it using list comprehension?

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You can't do it with a list comprehension because the result of a list comprehension is a list. You can do it with a dictionary comprehension, though: result = {key: value for dictionary in list_of_dictionaries for key, value in dictionary.items()}

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You could also just do: empty_dictionary = {} for dictionary in list_of_dictionaries: empty_dictionary.update(dictionary)