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How to create new dictionary from string?

I have a string: my_str = "abra cadabra"
I need to create a new dictionary , the keys in the dictionary are the letters in my_str and the value should be the amount of every letter.
For example:
my_str = "abra cadabra"
output >>
{'a': 5, 'b': 2, 'r': 2, 'c': 1, 'd': 1}
What i got is - {'a': 5, 'b': 5, 'r': 5, 'c': 5, 'd': 5}

Try this:
my_str = "abra cadabra"
my_set = set(my_str)
my_set.discard(" ")
my_dict = {}
for key in my_set:
my_dict[key] = my_str.count(key)
print(my_dict)

collections.Counter could be used for this.
Example:
from collections import Counter
my_str = "abra cadabra"
my_str = my_str.replace(" ", "")
c = Counter(list(my_str))
print('{}\n{}\n{}'.format(c, c.keys(), c.values()))
output:
Counter({'a': 5, 'b': 2, 'r': 2, 'c': 1, 'd': 1})
dict_keys(['a', 'b', 'r', 'c', 'd'])
dict_values([5, 2, 2, 1, 1])
You can iterate through the keys and values just like you would a normal dict.

Python How to add all values in a 2d Dictionary and return a single dictionary with summed values?

Part of the program I am developing has a 2D dict of length n.
Dictionary Example:
test_dict = {
0: {'A': 2, 'B': 1, 'C': 5},
1: {'A': 3, 'B': 1, 'C': 2},
2: {'A': 1, 'B': 1, 'C': 1},
3: {'A': 4, 'B': 2, 'C': 5}
}
All of the dictionaries have the same keys but different values. I need to sum all the values as to equal below.
I have tried to merge the dictionaries using the following:
new_dict = {}
for k, v in test_dict.items():
new_dict.setdefault(k, []).append(v)
I also tried using:
new_dict = {**test_dict[0], **test_dict[1], **test_dict[2], **test_dict[3]}
Unfortuntly I have not had any luck in getting the desired outcome.
Desired Outcome: outcome = {'A': 10, 'B': 5, 'C': 13}
How can I add all the values into a single dictionary?

Solution using pandas
Convert your dict to pandas.DataFrame and then do summation on columns and convert it back to dict.
import pandas as pd
df = pd.DataFrame.from_dict(test_dict, orient='index')
print(df.sum().to_dict())
Output:
{'A': 10, 'B': 5, 'C': 13}
Alternate solution
Use collections.Counter which allows you to add the values of same keys within dict
from collections import Counter
d = Counter()
for _,v in test_dict.items():
d.update(v)
print(d)

How to convert list of dictionaries to dictionaries

mylist = [{'a': 1, 'b': 2}, {'c': 3, 'd': 4}, {'e': 5, 'f': 6}]
i want it as
myDict ={'a': 1, 'b': 2,'c': 3, 'd': 4,'e': 5, 'f': 6}

You can make use of ChainMap.
from collections import ChainMap
myDict = dict(ChainMap(*mylist ))

This will take each dictionary and iterate through its key value pairs in for (k,v) in elem.items() part and assign them to a new dictionary.
mylist = [{'a': 1, 'b': 2}, {'c': 3, 'd': 4}, {'e': 5, 'f': 6}]
new_dict = {k:v for elem in mylist for (k,v) in elem.items()}
print new_dict
This will replace the duplicated keys.

I would create a new dictionary, iterate over the dictionaries in mylist, then iterate over the key/value pairs in that dictionary. From there, you can add each key/value pair to myDict.
mylist = [{'a': 1, 'b': 2}, {'c': 3, 'd': 4}, {'e': 5, 'f': 6}]
myDict = {}
for Dict in mylist:
for key in Dict:
myDict[key] = Dict[key]
print(myDict)

Store variables in dictionary for large data

I can print variables in python.
for h in jl1["results"]["attributes-list"]["volume-attributes"]:
state = str(h["volume-state-attributes"]["state"])
if aggr in h["volume-id-attributes"]["containing-aggregate-name"]:
if state == "online":
print(h["volume-id-attributes"]["owning-vserver-name"]),
print(' '),
print(h["volume-id-attributes"]["name"]),
print(' '),
print(h["volume-id-attributes"]["containing-aggregate-name"]),
print(' '),
print(h["volume-space-attributes"]["size-used"]
These print function returns for example 100 lines. Now I want to print only top 5 values based on filter of "size-used".
I am trying to take these values in dictionary and filter out top five values for "size-used" but not sure how to take them in dictionary.
Some thing like this
{'vserver': (u'rcdn9-c01-sm-prod',), 'usize': u'389120', 'vname': (u'nprd_root_m01',), 'aggr': (u'aggr1_n01',)}
Any other options like namedtuples is also appreciated.
Thanks

To get a list of dictionaries sorted by a certain key, use sorted. Say I have a list of dictionaries with a and b keys and want to sort them by the value of the b element:
my_dict_list = [{'a': 3, 'b': 1}, {'a': 1, 'b': 4}, {'a': 4, 'b': 4},
{'a': 2, 'b': 7}, {'a': 2, 'b': 4.3}, {'a': 2, 'b': 9}, ]
my_sorted_dict_list = sorted(my_dict_list, key=lambda element: element['b'], reverse=True)
# Reverse is set to True because by default it sorts from smallest to biggest; we want to reverse that
# Limit to five results
biggest_five_dicts = my_sorted_dict_list[:5]
print(biggest_five_dicts) # [{'a': 2, 'b': 9}, {'a': 2, 'b': 7}, {'a': 2, 'b': 4.3}, {'a': 1, 'b': 4}, {'a': 4, 'b': 4}]

heapq.nlargest is the obvious way to go here:
import heapq
interesting_dicts = ... filter to keep only the dicts you care about (e.g. online dicts) ...
for large in heapq.nlargest(5, interesting_dicts,
key=lambda d: d["volume-space-attributes"]["size-used"]):
print(...)

How to set a value by key for a dictionary in python using the map function

I know that I can set a key-value pair by using
dict[key] = value
but I have a very long list of dicts of the type
dict = [{a:1, b:2, c:3, d:4},
{a:2, b:3, c:4, d:5},
{a:5, b:7, c:3, d:9}]
and I'd like to do something along the lines of
dict = map(lambda x: x['d'] <- x['d'] -1, dict)
how would I go about this? (This is a very simplified example so I'm not really trying to just subtract a number from all items by a particular key)
expected output would be in this case and not the general case I'm looking for
[{a:1, b:2, c:3, d:3},
{a:2, b:3, c:4, d:4},
{a:5, b:7, c:3, d:8}]
EDIT: 2
I believe the following does not work - so any similar solution would be helpful:
dict = map(lambda x: x.update(d, x[d] - 1), dict)

dicts = [{'a':1, 'b':2, 'c':3, 'd':4},
{'a':2, 'b':3, 'c':4, 'd':5},
{'a':5, 'b':7, 'c':3, 'd':9}]
for d in dicts:
d['d'] -= 1
Output:
In [94]: dicts
Out[94]:
[{'d': 3, 'b': 2, 'c': 3, 'a': 1},
{'d': 4, 'b': 3, 'c': 4, 'a': 2},
{'d': 8, 'b': 7, 'c': 3, 'a': 5}]

how about this: as exactly you said
>>> dicts = [{'a':1, 'b':2, 'c':3, 'd':4},
{'a':2, 'b':3, 'c':4, 'd':5},
{'a':5, 'b':7, 'c':3, 'd':9}]
>>> map(lambda x:x.update([('d',x['d']-1)]),dicts)
[None, None, None]
>>> dicts
[{'a': 1, 'c': 3, 'b': 2, 'd': 3}, {'a': 2, 'c': 4, 'b': 3, 'd': 4}, {'a': 5, 'c': 3, 'b': 7, 'd': 8}]
update will update the dictionary with (key,value) pair. Returns None

map is a way of transforming an iterable to a list by performing the same operation on every item from the iterable. I don't think that's what you want to do here, and it has confused you.
On the face of it (although you haven't mentioned what the real operation is that you want to perform) a simple for is all that is necessary:
dict_list = [
{'a': 1, 'b': 2, 'c': 3, 'd': 4},
{'a': 2, 'b': 3, 'c': 4, 'd': 5},
{'a': 5, 'b': 7, 'c': 3, 'd': 9},
]
for d in dict_list:
d['d'] -= 1
print(d)
output
{'a': 1, 'b': 2, 'c': 3, 'd': 3}
{'a': 2, 'b': 3, 'c': 4, 'd': 4}
{'a': 5, 'b': 7, 'c': 3, 'd': 8}

Using dict.__setitem__ and temporary list (or any other collection typer) trick:
>>> dicts = [{'a':1, 'b':2, 'c':3, 'd':4},
... {'a':2, 'b':3, 'c':4, 'd':5},
... {'a':5, 'b':7, 'c':3, 'd':9}]
>>> map(lambda d: [d.__setitem__('d', d['d'] - 1), d][1], dicts)
[{'a': 1, 'c': 3, 'b': 2, 'd': 3},
{'a': 2, 'c': 4, 'b': 3, 'd': 4},
{'a': 5, 'c': 3, 'b': 7, 'd': 8}]
Using simple for loop is moe recommended way. Especially there's a side effect in the function.
BTW, don't use dict as a variable name. It will shadows builtin function/type dict.

How about this:
my_dict = {k: f(v) for k, v in my_dict.iteritems()}
where f is whatever function you want.