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Checking if value is present in the list of key-value pairs, if not present then add the new key-pair value from iterables (not overwriting same key)

If I used the not in it still appends the new key-pair value even if a specific value is already in the list.
dict1 = {'a': 0, 'a': 5, 'b': 1, 'c': 2}
list1 = [{'a': 0}] # This key-pair value is already in the list1 but
# still added from dict1.
new1 = []
new2 = []
for key, value in dict1.items():
if value not in list1:
new1.append(key)
new2.append(value)
new0 = {new1[i]: new2[i] for i in range(len(new1))}
list1.append(new0)
Desired output is:
list1 = [{'a': 0, 'a': 5, 'b': 1, 'c': 2}]
(As I dont want to overwrite the key/s)

As you do not provide example data, I have to make some guesses here. If I guessed incorrectly, please provide the required information.
You call .items on list1. A list does not have a items function. Instead, I suspect your list1 is actually a dictionary, which would like this for example:
list1 = {
"a":1,
"b":2,
"c":3
}
In your current loop, you check if the value is within list2. If list2 is actually a list, you're doing so correctly. However, based on your title I assume what you actually want to do is check if the key is in list2, and if not add the key:value to list2. You could not add a key:value pair to a list, so I assume that list2 shoudl also be a dictionary. You would be able to add them add as a tuple, but based on the title I assume that is not what you want.
If you actually want to add it as a key:value pair to a dictionary, you could do that as follows:
list2 = {}
for key, value in list1.items():
if key not in list2.keys(): # check if this key already exists
list2[key] = value # if not, add they key with the value
As list1 and list2 are not actually instances of list, but of dict I would recommend renaming your variables to avoid future confusion. Hope that helps!
EDIT after update in question
Your example data had a small mistake, as there were two a keys, meaning that the first {'a':0} would be overwritten within dict1 already.
dict1 = {'a': 0, 'b': 5, 'c': 1, 'd': 2}
list1 = [{'a': 0}]
As I understand it, you wish to check if the value is already containted within a list of dictionaries.
As such, we need to get all the values from these dictionaries.
As you do not want to overwrite any keys, it would need to be a list of dictionaries that each only have one key. As such, we can get each individual dictionary, get the keys. This returns an dict_keys object, which we can convert to a list. Since each dictionary within list1 always only has one key, we can just take the first from said lsit.
[list(x.values())[0] for x in list1]
Putting that within the loop we get
for key, value in dict1.items():
if not value in [list(x.values())[0] for x in list1]:
# If no dictionary exists within `list1` with this value
list1.append({key:value}) # then add it as a new dictionary
This would return
[{'a': 0}, {'b': 5}, {'c': 1}, {'d': 2}]
You could run this code again with a different dict1 and it would not overwrite keys within list1, for example with:
dict1 = {'a': 9}
the output would become
[{'a': 0}, {'b': 5}, {'c': 1}, {'d': 2}, {'a': 9}]

Access nested dictionary without loops [duplicate]

This question already has answers here:
Access nested dictionary items via a list of keys?
(20 answers)
Closed 4 years ago.
If I have a nested dictionary
mydict={'a1':{'b1':1}, 'a2':2}
and a list of indexes index = ['a1', 'b1'] leading to an inner value, is there a pythonic / one-liner way to get that value, i.e. without resorting to a verbose loop like:
d = mydict
for idx in index:
d = d[idx]
print(d)

You can use functools.reduce.
>>> from functools import reduce
>>> mydict = {'a1':{'b1':1}, 'a2':2}
>>> keys = ['a1', 'b1']
>>> reduce(dict.get, keys, mydict)
1
dict.get is a function that takes two arguments, the dict and a key (and another optional argument not relevant here). mydict is used as the initial value.
In case you ever need the intermediary results, use itertools.accumulate.
>>> from itertools import accumulate
>>> list(accumulate([mydict] + keys, dict.get))
[{'a1': {'b1': 1}, 'a2': 2}, {'b1': 1}, 1]
Unfortunately, the function does not take an optional initializer argument, so we are prepending mydict to keys.

How to create a dictionary out of a list of lists in python?

Let's suppose I have the following list made out of lists
list1 = [['a','b'],['a'],['b','c'],['c','d'],['b'], ['a','d']]
I am wondering if there is a way to convert every element of list1 in a dictionary where all the new dictionaries will use the same key. E.g: if ['a']
gets to be {'a':1}, and ['b'] gets to be {'b':2}, I would like for all keys a the value of 1 and for all keys b the value of 2. Therefore, when creating the dictionary of ['a','b'], I would like to turn into {'a':1, 'b':2}.
What I have found so far are ways to create a dictionary out of lists of lists but using the first element as the key and the rest of the list as the value:
Please note that's not what I am interested in.
The result I would want to obtain from list1 is something like:
dict_list1 = [{'a':1,'b':2}, {'a':1}, {'b':2,'c':3}, {'c':3,'d':4}, {'b':2}, {'a':1,'d':4}]
I am not that interested in the items being that numbers but in the numbers being the same for each different key.

You need to declare your mapping first:
mapping = dict(a=1, b=2, c=3, d=4)
Then, you can just use dict comprehension:
[{e: mapping[e] for e in li} for li in list1]
# [{'a': 1, 'b': 2}, {'a': 1}, {'b': 2, 'c': 3}, {'c': 3, 'd': 4}, {'b': 2}, {'a': 1, 'd': 4}]

Using chain and OrderedDict you can do auto mapping
from itertools import chain
from collections import OrderedDict
list1 = [['a','b'],['a'],['b','c'],['c','d'],['b'], ['a','d']]
# do flat list for auto index
flat_list = list(chain(*list1))
# remove duplicates
flat_list = list(OrderedDict.fromkeys(flat_list))
mapping = {x:flat_list.index(x)+1 for x in set(flat_list)}
[{e: mapping[e] for e in li} for li in list1]

Here a try with ord() also it will work for both capital and lower letters :
[{e: ord(e)%32 for e in li} for li in list1]

Multi-Line Dictionary to Single-Line Dictionary [duplicate]

This question already has answers here:
How do I merge two dictionaries in a single expression in Python?
(43 answers)
Closed 5 years ago.
How would you go about converting a multi-line dictionary into one dictionary?
For example, the current dictionary, if printed to the screen, is of the form:
{'keyword':'value'}
{'keyword':'value'}
{'keyword':'value'}
{'keyword':'value'}
...and so on for over 100 hundred lines. How do you convert this to the following form:
{'keyword':'value','keyword':'value','keyword':'value','keyword':'value'}

Assuming you are asking for multiple dictionaries (not multiple line dictionary) to one dictionary.
a = {1: 1, 2:2}
b = {2:2, 3:3}
c = {2:3}
{**a, **b, **c}
Out: {1: 1, 2: 3, 3: 3}

Assuming that your initial data is actually a list of dictionaries and your keys are unique accross all your dictionaries I would use something like -
example = [{'a':1}, {'b':2}, {'c':3}]
objOut = {}
for d in example:
for k,v in d.iteritems():
objOut[k] = v
OR
objIn = [{'a':1}, {'b':2}, {'c':3}]
objOut = {}
for d in objIn:
objOut.update(d)
print objOut

Given
dicts = [{'a':1}, {'b':2}, {'c':3, 'd':4}]
Do
{k:v for d in dicts for (k, v) in d.items()}
or
from itertools import chain
dict(chain(*map(dict.items, dicts)))
resulting in
{'a': 1, 'b': 2, 'c': 3, 'd': 4}

Merge multiple list of dict with same value in common key [closed]

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I have 4 list of dicts
list1 = [{'a':0,'b':23}, {'a':3,'b':77},{'a':1,'b':99}]
list2 = [{'a':1,'c':666},{'a':4,'c':546}]
list3 = [{'d':33,'a':3},{'d':1111,'a':4},{'d':76,'a':1},{'d':775,'a':0}]
list4 = [{'a':2,'e':12},{'a':4,'e':76}]
Every dict in the list has a common key 'a'. My requirement is 'a' key with same value in dict from all the list should be merged and if a particular key does not exist in the dicts while merging, then assign 0 for those keys or just omit those keys.
for eg. for key 'a' with value 1, from above example we have 2 dicts, one from list1 i.e {'a':0,'b':23} and one is list3, last dict i.e {'d':775,'a':0}, so 1st we have identified the dicts with same 'a' value, now need to merge these dicts
i.e {'a':0, 'b':23, 'c':0, 'd':775, 'e':0}, since both the dict didn't have 'c', 'c' is assigned as 0 here
I should get the output as:
[{'a':0,'b':23,'c':0,'d':775, 'e':0},{'a':1,'b':99,'c':666,'d':76,'e':0},{'a':2,'b':0,'c':0,'d':0,'e':12},{'a':3,'b':77,'c':0,'d':33,'e':0}, {'a':4,'b':0,'c':546,'d':1111,'e':76}]
usings minimum loops or list comprehension

If you want a more pythonic way:
from itertools import groupby
from pprint import pprint
from collections import ChainMap
a = [{'a':0,'b':23}, {'a':3,'b':77}, {'a':1,'b':99}]
b = [{'a':1,'c':666}, {'a':4,'c':546}]
c = [{'d':33,'a':3}, {'d':1111,'a':4}, {'d':76,'a':1}, {'d':775,'a':0}]
d = [{'a':2,'e':12}, {'a':4,'e':76}]
dict_list = a + b + c + d
# You just need to specify the key you want to use in the lambda function
# There's no need to declare the different key values previously
res = map(lambda dict_tuple: dict(ChainMap(*dict_tuple[1])),
groupby(sorted(dict_list,
key=lambda sub_dict: sub_dict["a"]),
key=lambda sub_dict: sub_dict["a"]))
pprint(list(res))
Outputs:
[{'a': 0, 'b': 23, 'd': 775},
{'a': 1, 'b': 99, 'c': 666, 'd': 76},
{'a': 2, 'e': 12},
{'a': 3, 'b': 77, 'd': 33},
{'a': 4, 'c': 546, 'd': 1111, 'e': 76}]
Edit (Improvement):
You can also use
from _operator import itemgetter
key=itemgetter("a")
instead of
key=lambda sub_dict: sub_dict["a"]
The version with itemgetter is much faster. Using the example you provided:
- Lambda: 0.037109375ms
- Itemgetter: 0.009033203125ms