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How can I add n times a value in a dictionary? [duplicate]

Can I use list comprehension syntax to create a dictionary?
For example, by iterating over pairs of keys and values:
d = {... for k, v in zip(keys, values)}

Use a dict comprehension (Python 2.7 and later):
{key: value for (key, value) in iterable}
Alternatively for simpler cases or earlier version of Python, use the dict constructor, e.g.:
pairs = [('a', 1), ('b', 2)]
dict(pairs) #=> {'a': 1, 'b': 2}
dict([(k, v+1) for k, v in pairs]) #=> {'a': 2, 'b': 3}
Given separate arrays of keys and values, use the dict constructor with zip:
keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values)) #=> {'a': 1, 'b': 2}
2) "zip'ped" from two separate iterables of keys/vals
dict(zip(list_of_keys, list_of_values))

In Python 3 and Python 2.7+, dictionary comprehensions look like the below:
d = {k:v for k, v in iterable}
For Python 2.6 or earlier, see fortran's answer.

In fact, you don't even need to iterate over the iterable if it already comprehends some kind of mapping, the dict constructor doing it graciously for you:
>>> ts = [(1, 2), (3, 4), (5, 6)]
>>> dict(ts)
{1: 2, 3: 4, 5: 6}
>>> gen = ((i, i+1) for i in range(1, 6, 2))
>>> gen
<generator object <genexpr> at 0xb7201c5c>
>>> dict(gen)
{1: 2, 3: 4, 5: 6}

Create a dictionary with list comprehension in Python
I like the Python list comprehension syntax.
Can it be used to create dictionaries too? For example, by iterating
over pairs of keys and values:
mydict = {(k,v) for (k,v) in blah blah blah}
You're looking for the phrase "dict comprehension" - it's actually:
mydict = {k: v for k, v in iterable}
Assuming blah blah blah is an iterable of two-tuples - you're so close. Let's create some "blahs" like that:
blahs = [('blah0', 'blah'), ('blah1', 'blah'), ('blah2', 'blah'), ('blah3', 'blah')]
Dict comprehension syntax:
Now the syntax here is the mapping part. What makes this a dict comprehension instead of a set comprehension (which is what your pseudo-code approximates) is the colon, : like below:
mydict = {k: v for k, v in blahs}
And we see that it worked, and should retain insertion order as-of Python 3.7:
>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah2': 'blah', 'blah3': 'blah'}
In Python 2 and up to 3.6, order was not guaranteed:
>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah3': 'blah', 'blah2': 'blah'}
Adding a Filter:
All comprehensions feature a mapping component and a filtering component that you can provide with arbitrary expressions.
So you can add a filter part to the end:
>>> mydict = {k: v for k, v in blahs if not int(k[-1]) % 2}
>>> mydict
{'blah0': 'blah', 'blah2': 'blah'}
Here we are just testing for if the last character is divisible by 2 to filter out data before mapping the keys and values.

In Python 2.7, it goes like:
>>> list1, list2 = ['a', 'b', 'c'], [1,2,3]
>>> dict( zip( list1, list2))
{'a': 1, 'c': 3, 'b': 2}
Zip them!

Python version >= 2.7, do the below:
d = {i: True for i in [1,2,3]}
Python version < 2.7(RIP, 3 July 2010 - 31 December 2019), do the below:
d = dict((i,True) for i in [1,2,3])

To add onto #fortran's answer, if you want to iterate over a list of keys key_list as well as a list of values value_list:
d = dict((key, value) for (key, value) in zip(key_list, value_list))
or
d = {(key, value) for (key, value) in zip(key_list, value_list)}

Just to throw in another example. Imagine you have the following list:
nums = [4,2,2,1,3]
and you want to turn it into a dict where the key is the index and value is the element in the list. You can do so with the following line of code:
{index:nums[index] for index in range(0,len(nums))}

Here is another example of dictionary creation using dict comprehension:
What i am tring to do here is to create a alphabet dictionary where each pair; is the english letter and its corresponding position in english alphabet
>>> import string
>>> dict1 = {value: (int(key) + 1) for key, value in
enumerate(list(string.ascii_lowercase))}
>>> dict1
{'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4, 'g': 7, 'f': 6, 'i': 9, 'h': 8,
'k': 11, 'j': 10, 'm': 13, 'l': 12, 'o': 15, 'n': 14, 'q': 17, 'p': 16, 's':
19, 'r': 18, 'u': 21, 't': 20, 'w': 23, 'v': 22, 'y': 25, 'x': 24, 'z': 26}
>>>
Notice the use of enumerate here to get a list of alphabets and their indexes in the list and swapping the alphabets and indices to generate the key value pair for dictionary
Hope it gives a good idea of dictionary comp to you and encourages you to use it more often to make your code compact

This code will create dictionary using list comprehension for multiple lists with different values that can be used for pd.DataFrame()
#Multiple lists
model=['A', 'B', 'C', 'D']
launched=[1983,1984,1984,1984]
discontinued=[1986, 1985, 1984, 1986]
#Dictionary with list comprehension
keys=['model','launched','discontinued']
vals=[model, launched,discontinued]
data = {key:vals[n] for n, key in enumerate(keys)}
#Convert dict to dataframe
df=pd.DataFrame(data)
display(df)
enumerate will pass n to vals to match each key with its list

Try this,
def get_dic_from_two_lists(keys, values):
return { keys[i] : values[i] for i in range(len(keys)) }
Assume we have two lists country and capital
country = ['India', 'Pakistan', 'China']
capital = ['New Delhi', 'Islamabad', 'Beijing']
Then create dictionary from the two lists:
print get_dic_from_two_lists(country, capital)
The output is like this,
{'Pakistan': 'Islamabad', 'China': 'Beijing', 'India': 'New Delhi'}

Adding to #Ekhtiar answer, if you want to make look up dict from list, you can use this:
names = ['a', 'b', 'd', 'f', 'c']
names_to_id = {v:k for k, v in enumerate(names)}
# {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'f': 4}
Or in rare case that you want to filter duplicate, use set first (best in list of number):
names = ['a', 'b', 'd', 'f', 'd', 'c']
sorted_list = list(set(names))
sorted_list.sort()
names_to_id = {v:k for k, v in enumerate(sorted_list)}
# {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'f': 4}
names = [1,2,5,5,6,2,1]
names_to_id = {v:k for k, v in enumerate(set(names))}
# {1: 0, 2: 1, 5: 2, 6: 3}

>>> {k: v**3 for (k, v) in zip(string.ascii_lowercase, range(26))}
Python supports dict comprehensions, which allow you to express the creation of dictionaries at runtime using a similarly concise syntax.
A dictionary comprehension takes the form {key: value for (key, value) in iterable}. This syntax was introduced in Python 3 and backported as far as Python 2.7, so you should be able to use it regardless of which version of Python you have installed.
A canonical example is taking two lists and creating a dictionary where the item at each position in the first list becomes a key and the item at the corresponding position in the second list becomes the value.
The zip function used inside this comprehension returns an iterator of tuples, where each element in the tuple is taken from the same position in each of the input iterables. In the example above, the returned iterator contains the tuples (“a”, 1), (“b”, 2), etc.
Output:
{'i': 512, 'e': 64, 'o': 2744, 'h': 343, 'l': 1331, 's': 5832, 'b': 1, 'w': 10648, 'c': 8, 'x': 12167, 'y': 13824, 't': 6859, 'p': 3375, 'd': 27, 'j': 729, 'a': 0, 'z': 15625, 'f': 125, 'q': 4096, 'u': 8000, 'n': 2197, 'm': 1728, 'r': 4913, 'k': 1000, 'g': 216, 'v': 9261}

Yes, it's possible. In python, Comprehension can be used in List, Set, Dictionary, etc.
You can write it this way
mydict = {k:v for (k,v) in blah}
Another detailed example of Dictionary Comprehension with the Conditional Statement and Loop:
parents = [father, mother]
parents = {parent:1 - P["mutation"] if parent in two_genes else 0.5 if parent in one_gene else P["mutation"] for parent in parents}

You can create a new dict for each pair and merge it with the previous dict:
reduce(lambda p, q: {**p, **{q[0]: q[1]}}, bla bla bla, {})
Obviously this approaches requires reduce from functools.

Assuming blah blah blah is a two-tuples list:
Let's see two methods:
# method 1
>>> lst = [('a', 2), ('b', 4), ('c', 6)]
>>> dict(lst)
{'a': 2, 'b': 4, 'c': 6}
# method 2
>>> lst = [('a', 2), ('b', 4), ('c', 6)]
>>> d = {k:v for k, v in lst}
>>> d
{'a': 2, 'b': 4, 'c': 6}

this approach uses iteration over the given date using a for loop.
Syntax: {key: value for (key, value) in data}
Eg:
# create a list comprehension with country and code:
Country_code = [('China', 86), ('USA', 1),
('Ghana', 233), ('Uk', 44)]
# use iterable method to show results
{key: value for (key, value) in Country_code}

Appending a dictionary of values to a dictionary of lists with same keys

I am given a dictionary of values and a dictionary of lists
dict_v = {'A': 2, 'B': 3, 'C': 4}
&
dict_l = {'A': [1,3,4], 'B': [8,5,2], 'C': [4,6,2]}
I am trying to append the values of dict_v to the values of dict_l.
So I want it to look like this:
{'A': [1,3,4,2], 'B': [8,5,2,3], 'C': [4,6,2,4]}
Ive tried the following code:
for key in dict_l:
if key in dict_v:
dict_v[key]=dict_l[key]+dict_v[key]
else:
dict_l[key]=stock_prices[key]
print(dict_v)
it's giving me an error message

You can append to a list in this sense like this
dict_v = {'A': 2, 'B': 3, 'C': 4}
dict_l = {'A': [1,3,4], 'B': [8,5,2], 'C': [4,6,2]}
for key in dict_l:
if key in dict_v:
dict_l[key] += [dict_v[key]]
print(dict_l)
The change is that you are appending the value dict_v[key] as a list [dict_v[key]] and appending it to the entries that are already in dict_l[key] using +=. This way you can also append multiple values as a list to an already existing list.

because you are adding to a list you should user the append() method to add to list
dict_v = {'A': 2, 'B': 3, 'C': 4}
dict_l = {'A': [1,3,4], 'B': [8,5,2], 'C': [4,6,2]}
for key in dict_v.keys():
dict_l[key].append(dict_v[key])
print(dict_l)

Here is my answer:
dict_v = {'A': 2, 'B': 3, 'C': 4}
dict_l = {'A': [1,3,4], 'B': [8,5,2], 'C': [4,6,2]}
for key in dict_v:
if key in dict_l:
temp = dict_l[key]
temp.append(dict_v[key])
dict_l[key] = temp
else:
dict_l[key] = stock_prices[key]
print(dict_l)
So instead of for key in dict_l, I did for key in dict_v just for the sake of logic. I also appended key value from dict_v to dict_l, but in your code you appended the key value from dict_l to dict_v, and I changed that. Hope it helps!

Merge values of dictionary into one if keys match condition

I have a dictionary called, say, tempDict which is filled as such:
tempDict = {'a': 100, 'b': 200, 'c': 120, 'b_ext1': 4, 'd': 1021, 'a_ext1': 21, 'f_ext1': 12}
In my setup, I need to loop over the keys and if for any key that has a '_ext1' suffix, I want to rewrite or create a new dictionary keeping an unaltered key (preferably without the 'ext1') but with the values merged.
i.e.:
newDict = {'a': 121, 'b': 204, 'c': 120, 'd': 1021, 'f_ext1':12}
Notice, the last entry in the dictionary should be unchanged as there is no 'f' unsuffixed with '_ext1'
The values themselves won't be integers, however the operation is similar.
Does anyone have any ideas?

newDict = {}
for k in tempDict:
if k.endswith("_ext1") and k[:-5] in tempDict:
newDict[k[:-5]] = newDict.get(k[:-5],0)+tempDict[k]
else:
newDict[k] = newDict.get(k,0)+tempDict[k]

Iterate over the keys (items) in the sorted order and sum or add values. This works because keys like "a", "f" sort before "a_ext1", "f_ext1":
>>> d = {}
>>> for k, v in sorted(tempDict.items()):
... if k[0] in d:
... d[k[0]] += v
... else:
... d[k] = v
...
>>> d
{'a': 121, 'c': 120, 'b': 204, 'd': 1021, 'f_ext1': 12}

Pythonic way of finding duplicate maps in a list while ignoring certain keys, and combining the duplicate maps to make a new list

I want to write a code which takes the following inputs:
list (list of maps)
request_keys (list of strings)
operation (add,substract,multiply,concat)
The code would look at the list for the maps having the same value for all keys except the keys given in request_keys. Upon finding two maps for which the value in the search keys match, the code would do the operation (add,multiple,substract,concat) on the two maps and combine them into one map. This combination map would basically replace the other two maps.
i have written the following peice of code to do this. The code only does add operation. It can be extended to make the other operations
In [83]: list
Out[83]:
[{'a': 2, 'b': 3, 'c': 10},
{'a': 2, 'b': 3, 'c': 3},
{'a': 2, 'b': 4, 'c': 4},
{'a': 2, 'b': 3, 'c': 2},
{'a': 2, 'b': 3, 'c': 3}]
In [84]: %cpaste
Pasting code; enter '--' alone on the line to stop or use Ctrl-D.
:def func(list,request_keys):
: new_list = []
: found_indexes = []
: for i in range(0,len(list)):
: new_item = list[i]
: if i in found_indexes:
: continue
: for j in range(0,len(list)):
: if i != j and {k: v for k,v in list[i].iteritems() if k not in request_keys} == {k: v for k,v in list[j].iteritems() if k not in request_keys}:
: found_indexes.append(j)
: for request_key in request_keys:
: new_item[request_key] += list[j][request_key]
: new_list.append(new_item)
: return new_list
:--
In [85]: func(list,['c'])
Out[85]: [{'a': 2, 'b': 3, 'c': 18}, {'a': 2, 'b': 4, 'c': 4}]
In [86]:
What i want to know is, is there a faster, more memory efficient, cleaner and a more pythonic way of doing the same?
Thank you

You manually generate all the combinations and then compare each of those combinations. This is pretty wasteful. Instead, I suggest grouping the dictionaries in another dictionary by their matching keys, then adding the "same" dictionaries. Also, you forgot the operator parameter.
import collections, operator, functools
def func(lst, request_keys, op=operator.add):
matching_dicts = collections.defaultdict(list)
for d in lst:
key = tuple(sorted(((k, d[k]) for k in d if k not in request_keys)))
matching_dicts[key].append(d)
for group in matching_dicts.values():
merged = dict(group[0])
merged.update({key: functools.reduce(op, (g[key] for g in group))
for key in request_keys})
yield merged
What this does: First, it creates a dictionary, mapping the key-value pairs that have to be equal for two dictionaries to match to all those dictionaries that have those key-value pairs. Then it iterates the dicts from those groups, using one of that group as a prototype and updating it with the sum (or product, or whatever, depending on the operator) of the all the dicts in that group for the required_keys.
Note that this returns a generator. If you want a list, just call it like list(func(...)), or accumulate the merged dicts in a list and return that list.

from itertools import groupby
from operator import itemgetter
def mergeDic(inputData, request_keys):
keys = inputData[0].keys()
comparedKeys = [item for item in keys if item not in request_keys]
grouper = itemgetter(*comparedKeys)
result = []
for key, grp in groupby(sorted(inputData, key = grouper), grouper):
temp_dict = dict(zip(comparedKeys, key))
for request_key in request_keys:
temp_dict[request_key] = sum(item[request_key] for item in grp)
result.append(temp_dict)
return result
inputData = [{'a': 2, 'b': 3, 'c': 10},
{'a': 2, 'b': 3, 'c': 3},
{'a': 2, 'b': 4, 'c': 4},
{'a': 2, 'b': 3, 'c': 2},
{'a': 2, 'b': 3, 'c': 3}]
from pprint import pprint
pprint(mergeDic(inputData,['c']))

Adding more values on existing python dictionary key

I am new to python and i am stuck while making a dictionary.. please help :)
This is what I am starting with :
dict = {}
dict['a']={'ra':7, 'dec':8}
dict['b']={'ra':3, 'dec':5}
Everything perfect till now. I get :
In [93]: dict
Out[93]: {'a': {'dec':8 , 'ra': 7}, 'b': {'dec': 5, 'ra': 3}}
But now, if I want to add more things to key 'a' and i do :
dict['a']={'dist':12}
Then it erases the previous information of 'a' and what i get now is :
In [93]: dict
Out[93]: {'a': {'dist':12}, 'b': {'dec': 5, 'ra': 3}}
What i get want to have is :
In [93]: dict
Out[93]: {'a': {'dec':8 , 'ra': 7, 'dist':12}, 'b': {'dec': 5, 'ra': 3}}
Can someone please help??

>>> d = {}
>>> d['a'] = {'ra':7, 'dec':8}
>>> d['b'] = {'ra':3, 'dec':5}
>>> d['a']['dist'] = 12
>>> d
{'a': {'dec': 8, 'dist': 12, 'ra': 7}, 'b': {'dec': 5, 'ra': 3}}
If you want to update dictionary from another dictionary, use update():
Update the dictionary with the key/value pairs from other, overwriting
existing keys.
>>> d = {}
>>> d['a'] = {'ra':7, 'dec':8}
>>> d['b'] = {'ra':3, 'dec':5}
>>> d['a'].update({'dist': 12})
>>> d
{'a': {'dec': 8, 'dist': 12, 'ra': 7}, 'b': {'dec': 5, 'ra': 3}}
Also, don't use dict as a variable name - it shadows built-in dict type. See what can possibly happen:
>>> dict(one=1)
{'one': 1}
>>> dict = {}
>>> dict(one=1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'dict' object is not callable

Do this:
dict['a']['dist'] = 12

Try this:
dict['a'].update( {'dist': 12} )

Instead of assigning {'dist':12} to dict['a'], use the update method.
dict['a'].update( {'dist':12} )
This has the advantage of not needing to "break apart" the new dictionary to find which key(s) to insert into the target. Consider:
a = build_some_dictionary()
for k in a:
dict['a'] = a[k]
vs.
dict['a'].update(a)