I have a list and I want to convert to a specific dict dict_a
Input:
a = ['AAA: key1', 'value1', 'value2', 'AAA: key2', 'value3', 'value4', 'value5', 'AAA: key3', 'value6', 'value7']
Output expected:
dict_a = {'key1': [value1, value2], 'key2': [value3, value4, value5], 'key3': [value6, value7]}
My attempt:
for elem in a:
if a.startswith(AAA:):
d_a = {elem}
This is one way using collections.defaultdict:
from collections import defaultdict
a = ['AAA: key1', ' value1', ' value2', 'AAA: key2', ' value3', ' value4', ' value5', 'AAA: key3', ' value6', ' value7']
d = defaultdict(list)
for x in a:
if not x.startswith('AAA'):
d[c].append(x.strip())
else:
c = x.split(': ')[1]
print(d)
# defaultdict(<class 'list'>, {'key1': ['value1', 'value2'], 'key2': ['value3', 'value4', 'value5'], 'key3': ['value6', 'value7']})
I like to use a generator for problems like this, accumulating a result unitl we see the next result start, then yielding the result we've collected.
def key_val_gen(a):
key = None
vals = []
for item in a:
if item.startswith('AAA: '):
if key:
yield key, vals
key = item.split(maxsplit=1)[1]
vals = []
else:
vals.append(item)
if key:
yield key, vals
a = ['AAA: key1', 'value1', 'value2', 'AAA: key2', 'value3', 'value4', 'value5', 'AAA: key3', 'value6', 'value7']
print(dict(key_val_gen(a)))
# {'key1': ['value1', 'value2'], 'key2': ['value3', 'value4', 'value5'], 'key3': ['value6', 'value7']}
from collections import defaultdict
result = defaultdict(list)
key = None
a = ['AAA: key1', ' value1', ' value2', 'AAA: key2', ' value3', ' value4', ' value5', 'AAA: key3', ' value6', ' value7']
for elem in a:
if elem.startswith('AAA:'):
key = elem.split(':')[1].strip()
elif key:
result[key].append(elem.strip())
print(result)
You could also use itertools.groupby to group items in "is a key" and "is a value" and then combine consecutive elements by using next on the same iterator:
>>> from itertools import groupby
>>> groups = (next(g).lstrip("AAA: ") if k else list(g)
... for k, g in groupby(a, key=lambda x: x.startswith("AAA: ")))
...
>>> {g: next(groups) for g in groups}
{'key1': ['value1', 'value2'],
'key2': ['value3', 'value4', 'value5'],
'key3': ['value6', 'value7']}
Related
Very similar to this question but with an added caveat.
I have two lists identical in length. One contains my keys, and the other contains a list of dictionaries belonging to said keys. Example below (yes the Nones are intentional)
keys = ['key1', 'key2, 'key3']
vals = [[{'subkey1': 'val1', 'subkey2': 'val2'}], [{'subkey1: 'val2'},None],[{'subkey1':'val3'}, {'subkey3':'val1}, None, None]]
What I'd like to do is match each dictionary in each list to a key to create a nested dict.
So something like below:
final = {'key1':{'subkey1': 'val1', 'subkey2': 'val2'}, 'key2':{'subkey1': 'val2'}, 'key3':{'subkey1':'val3'}, {'subkey3':'val1'}}
I know we can use dictionary comprehension to do this. I created a dictionary of empty dicts.
s = {}
for i in keys:
for v in vals:
s[i] = {}
break
Then I wrote the following to iterate through the list of dicts I'd like to pair up accounting for the non dictionary values in my list.
x = 0
while x < len(keys):
for i in keys:
for element in vals[x]:
if isinstance(element, dict) == True:
s[str(i)].append(element)
else:
print('no')
x=x+1
However when I run this, I get AttributeError: 'NoneType' object has no attribute 'append'
How can I append to a dictionary using this for loop?
For the less readable dictionary comprehension solution
keys = ['key1', 'key2', 'key3']
vals = [
[{'subkey1': 'val1', 'subkey2': 'val2'}],
[{'subkey1': 'val2'}, None],
[{'subkey1': 'val3'}, {'subkey3':'val1'}, None, None]
]
s = {
k: {
sk: sv for d in (x for x in v if x is not None) for sk, sv in d.items()
} for k, v in zip(keys, vals)
}
Gives
{'key1': {'subkey1': 'val1', 'subkey2': 'val2'},
'key2': {'subkey1': 'val2'},
'key3': {'subkey1': 'val3', 'subkey3': 'val1'}}
This can be done fairly easily using a defaultdict:
from collections import defaultdict
keys = ['key1', 'key2', 'key3']
vals = [
[{'subkey1': 'val1', 'subkey2': 'val2'}],
[{'subkey1': 'val2'},None],
[{'subkey1':'val3'}, {'subkey3':'val1'}, None, None]
]
final = defaultdict(dict)
for idx, key in enumerate(keys):
for d in vals[idx]:
if d is not None:
final[key].update(d)
Output:
defaultdict(<class 'dict'>, {
'key1': {'subkey1': 'val1', 'subkey2': 'val2'},
'key2': {'subkey1': 'val2'},
'key3': {'subkey1': 'val3', 'subkey3': 'val1'}
})
can't use append on a dict
try this maybe?
keys = ['key1', 'key2', 'key3']
vals = [[{'subkey1': 'val1', 'subkey2': 'val2'}], [{'subkey1': 'val2'},None],[{'subkey1':'val3'}, {'subkey3':'val1'}, None, None]]
s = {}
for i in keys:
s[i] = {}
for i, key in enumerate(keys):
for element in vals[i]:
if isinstance(element, dict) == True:
s[str(key)].update(element)
print(s)
Output:
{'key1': {'subkey1': 'val1', 'subkey2': 'val2'}, 'key2': {'subkey1': 'val2'}, 'key3': {'subkey1': 'val3', 'subkey3': 'val1'}}
I want to create a dictionary from a string that have key=value
s = "key1=value1 key2=value2 key3=value3"
print({r.split("=") for r in s})
Is it possible using dictionary comprehension? If yes, how?
You can first split on whitespace, then split on '='
>>> dict(tuple(i.split('=')) for i in s.split())
{'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}
You could use map:
>>> s = "key1=value1 key2=value2 key3=value3"
>>> d = {k: v for k, v in map(lambda i: i.split('='), s.split())}
>>> {'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}
I have list of dictionary
dictio =[{'key1':'value1'}, {'key1':'value2'}, {'key1':'value1'}, {'key2':'value4'}, {'key2':'value5'}]
from collections import defaultdict
result = defaultdict(list)
for subd in dictio:
for k, v in subd.items():
result[k].append(v)
result
My output
While appending 'value1' appending to 'key1' which is not required
defaultdict(list,
{'key1': ['value1', 'value2', 'value1'],
'key2': ['value4', 'value5']})
My Expected
`defaultdict(list,
{'key1': ['value1', 'value2'],
'key2': ['value4', 'value5']})`
You can turn your list of dict into a set and then in a list again, a set would remove all duplicates.
dictio = list(set([{'key1':'value1'}, {'key1':'value2'}, {'key1':'value1'}, {'key2':'value4'}, {'key2':'value5'}]))
You can achieve that very simply, just check if the element is already in:
dictio =[{'key1':'value1'}, {'key1':'value2'}, {'key1':'value1'}, {'key2':'value4'}, {'key2':'value5'}]
from collections import defaultdict
result = defaultdict(list)
for subd in dictio:
for k, v in subd.items():
if v not in result[k]:
result[k].append(v)
Instead of list use set
Ex:
dictio =[{'key1':'value1'}, {'key1':'value2'}, {'key1':'value1'}, {'key2':'value4'}, {'key2':'value5'}]
from collections import defaultdict
result = defaultdict(set)
for subd in dictio:
for k, v in subd.items():
result[k].add(v)
result
# --> defaultdict(<class 'set'>, {'key1': {'value2', 'value1'}, 'key2': {'value4', 'value5'}})
Try this example from https://www.w3schools.com/python/python_howto_remove_duplicates.asp
mylist = ["a", "b", "a", "c", "c"]
mylist = list(dict.fromkeys(mylist))
print(mylist)
I'm trying to invert items of a dictionary containing strings as key and string list as value:
dico = {'key1': [],
'key2': [],
'key3': ['value1', 'value1'],
'key4': ['value2', 'value2'],
'key5': ['value3'],
'key6': ['value1', 'value2', 'value3']}
new_dict = {}
for key, values in dico.items():
if values:
for value in values:
try:
if key not in new_dict[value]:
new_dict[value].append(key)
except KeyError:
new_dict[values[0]] = list(key)
else:
print('ERROR')
Here's the result expected:
#Expected
new_dict = {'value1': ['key3', 'key6'],
'value2': ['key4', 'key6'],
'value3': ['key5', 'key6']}
#Reality
new_dict = {'value1': ["k", "e", "y", "3", "k", "e", "y", "6"],
'value2': ["k", "e", "y", "4", "k", "e", "y", "6"],
'value3': ["k", "e", "y", "5", "k", "e", "y", "6"]}
I noticed if I change that:
new_dict[values[0]] = list(key)
by that:
new_dict[values[0]] = []
new_dict[values[0]].append(key)
It actually works but is there another way to do it in one line ?
You are turning your keys to lists:
new_dict[values[0]] = list(key)
That'll produce a list with individual characters. Use a list literal instead:
new_dict[values[0]] = [key]
You can use the dict.setdefault() method to handle missing keys in new_dict to simplify your code. It looks like you want to produce sets instead; sets track unique values and saves you having to do explicit tests for duplicates.
for key, values in dico.items():
for value in values:
new_dict.setdefault(value, set()).add(key)
You can always turn those sets back to lists afterwards:
new_dict = {key: list(values) for key, values in new_dict.items()}
Demo:
>>> dico = {'key1': [],
... 'key2': [],
... 'key3': ['value1', 'value1'],
... 'key4': ['value2', 'value2'],
... 'key5': ['value3'],
... 'key6': ['value1', 'value2', 'value3']}
>>> new_dict = {}
>>> for key, values in dico.items():
... for value in values:
... new_dict.setdefault(value, set()).add(key)
...
>>> new_dict
{'value3': set(['key6', 'key5']), 'value2': set(['key6', 'key4']), 'value1': set(['key3', 'key6'])}
>>> {key: list(values) for key, values in new_dict.items()}
{'value3': ['key6', 'key5'], 'value2': ['key6', 'key4'], 'value1': ['key3', 'key6']}
Iterate every item from main dico dictionary.
Check if value is present or not.
Iterate every item from value.
Use set method to remove duplicate values.
Add to new_dict dictionary where value is key and key is list value.
code:
dico = {'key1': [],
'key2': [],
'key3': ['value1', 'value1'],
'key4': ['value2', 'value2'],
'key5': ['value3'],
'key6': ['value1', 'value2', 'value3']}
new_dict = {}
for key, values in dico.items():
if values:
for value in set(values):
try:
new_dict[value].append(key)
except:
new_dict[value] = [key]
import pprint
pprint.pprint(new_dict)
Output:
$ python test.py
{'value1': ['key3', 'key6'],
'value2': ['key6', 'key4'],
'value3': ['key6', 'key5']}
I have this list:
single = ['key1', 'value1', 'key2', 'value2', 'key3', 'value3']
What's the best way to create a dictionary from this?
Thanks.
>>> single = ['key1', 'value1', 'key2', 'value2', 'key3', 'value3']
>>> dict(zip(single[::2], single[1::2]))
{'key3': 'value3', 'key2': 'value2', 'key1': 'value1'}
Similar to SilentGhost's solution, without building temporary lists:
>>> from itertools import izip
>>> single = ['key1', 'value1', 'key2', 'value2', 'key3', 'value3']
>>> si = iter(single)
>>> dict(izip(si, si))
{'key3': 'value3', 'key2': 'value2', 'key1': 'value1'}
This is the simplest I guess. You will see more wizardry in solution here using list comprehension etc
dictObj = {}
for x in range(0, len(single), 2):
dictObj[single[x]] = single[x+1]
Output:
>>> single = ['key1', 'value1', 'key2', 'value2', 'key3', 'value3']
>>> dictObj = {}
>>> for x in range(0, len(single), 2):
... dictObj[single[x]] = single[x+1]
...
>>>
>>> dictObj
{'key3': 'value3', 'key2': 'value2', 'key1': 'value1'}
>>>
L = ['key1', 'value1', 'key2', 'value2', 'key3', 'value3']
d = dict(L[n:n+2] for n in xrange(0, len(L), 2))
>>> single = ['key', 'value', 'key2', 'value2', 'key3', 'value3']
>>> dict(zip(*[iter(single)]*2))
{'key3': 'value3', 'key2': 'value2', 'key': 'value'}
Probably not the most readable version though ;)
You haven't specified any criteria for "best". If you want understandability, simplicity, easily modified to check for duplicates and odd number of inputs, works with any iterable (in case you can't find out the length in advance), NO EXTRA MEMORY USED, ... try this:
def load_dict(iterable):
d = {}
pair = False
for item in iterable:
if pair:
# insert duplicate check here
d[key] = item
else:
key = item
pair = not pair
if pair:
grumble_about_odd_length(key)
return d