I have a list called rest which contains many dictionaries in a list which is in the format
rest = [{'a':'b','c':'d','e':'f'}, {'a':'g','c':'h','e':'i}, {'a':'j','c':'k','e':'l'}]
Can I get an output as below where I have new as a key inside a dictionary for all the key-value pair except the first key-value pair
output = [{'a':'b','new':{'c':'d','e':'f'}},{'a':'g','new':{'c':'h','e':'i'}},{'a':'j','new':{'c':'k','e':'l'}}]
Is it possible?
You can use the syntax first, *remainder to extract the relevant parts and then create a new dict from them:
def convert(d):
first, *remainder = d.items()
return dict([first, ('new', dict(remainder))])
Then to convert each of the dicts:
output = [convert(d) for d in rest]
Note that this syntax was introduced in Python 3.0 and dictionaries are unordered before Python 3.6 (i.e. the first item is not determined).
Related
If I have the following list of ids:
id_numbers = ['08a592a1-5a40-46b7-b9b3-b3a1e6bfdb9f', 'a31017f7-13b4-401c-81ca-fd57042bc181']
I can add these id numbers as values of a defined key "id_numbers" in a new dictionary using the following:
catalogue_ids = {"id_numbers": []}
for item in id_numbers:
catalogue_ids["id_numbers"].append(item)
I was wondering if there is a dictionary comprehension way of doing the same thing?
catalogue_ids = {"id_numbers": [item for item in id_numbers]}
There's no need to use list/dictionary comprehensions. You can simply assign the id_numbers variable to the dictionary element.
catalogue_ids = {"id_numbers": id_numbers}
I have a list of strings. Each string has the same length/number of characters in the format
xyzw01.ext or xyzv02.ext, etc.
For example
list 1: ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
list 2: ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
I would like from these lists to build new lists with only the strings with highest number.
So from list 1 I would like to get
['ADEJ01.ext','ABCJ02.ext','CDEJ03.ext']
while for list 2 I would like to get the same list since all numbers are 01.
Is there a "simple" way of achieving this?
You can use defaultdict and max
from collections import defaultdict
def fun(lst):
res = defaultdict(list)
for x in lst:
res[x[:4]].append(x)
return [max(res[x], key=lambda x: x[4:6]) for x in res]
lst = ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
lst2 = ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
print(fun(lst))
print(fun(lst2))
Output:
['ABCJ02.ext', 'CDEJ03.ext', 'ADEJ01.ext']
['ABCJ01.ext', 'ADEJ01.ext', 'CDEJ01.ext', 'RPNJ01.ext', 'PLEJ01.ext']
The easiest way is probably to use an intermediate data structure, like a dict - sort the list items into buckets based on the first part of their names, and then take the maximum number for each bucket. We can just use the built-in max() without a key, since as-given lexicographic sorting works to find the largest. If that's not sufficient, you could use more regex to take the number out of the item and use it as the key instead.
import re
def filter_list(lst):
prefixes = {}
for item in lst:
# use regex to isolate the non-numeric characters at the start of the string
prefix = re.match(r'^([^0-9]*)', item).group(1)
# make a bucket based on each prefix, and put the item in it
prefixes.setdefault(prefix, [])
prefixes[prefix].append(item)
# make a list comprehension taking the maximum item from each bucket
return [max(value) for value in prefixes.values()]
>>> a = ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
>>> b = ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
>>> filter_list(a)
['ABCJ02.ext', 'CDEJ03.ext', 'ADEJ01.ext']
>>> filter_list(b)
['ABCJ01.ext', 'ADEJ01.ext', 'CDEJ01.ext', 'RPNJ01.ext', 'PLEJ01.ext']
In python 3.7+, this should preserve the order of list from the first occurrence of each prefix (i.e. CDEJ03.ext will precede ADEJ01.ext in the output because CDEJ02.ext precedes it in the input).
To get the output in the exact same order as the original list, behavior, you'd want to explicitly reassign the key instead of using .setdefault(), perhaps with a pattern like prefixes[prefix] = prefixes[prefix] if prefix in prefixes else [].
Maybe it is ordinary issue regarding iterating thru a dict. Please find below imovel.txt file, whose content is as follows:
{'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
As you can see this is not a ordinary dictionary, with a key value pair; but a key with a list as key and another list as value
My code is:
#/usr/bin/python
def load_dict_from_file():
f = open('../txt/imovel.txt','r')
data=f.read()
f.close()
return eval(data)
thisdict = load_dict_from_file()
for key,value in thisdict.items():
print(value)
and yields :
['primeiro', 'segundo', 'terceiro'] ['101', '201', '301']
I would like to print a key,value pair like
{'primeiro':'101, 'segundo':'201', 'terceiro':'301'}
Given such txt file above, is it possible?
You should use the builtin json module to parse but either way, you'll still have the same structure.
There are a few things you can do.
If you know both of the base key names('Andar' and 'Apto') you can do it as a one line dict comprehension by zipping the values together.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# One line dict comprehension
newdict = {key: value for key, value in zip(thisdict['Andar'], thisdict['Apto'])}
print(newdict)
If you don't know the names of the keys, you could call next on an iterator assuming they're the first 2 lists in your structure.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# create an iterator of the values since the keys are meaningless here
iterator = iter(thisdict.values())
# the first group of values are the keys
keys = next(iterator, None)
# and the second are the values
values = next(iterator, None)
# zip them together and have dict do the work for you
newdict = dict(zip(keys, values))
print(newdict)
As other folks have noted, that looks like JSON, and it'd probably be easier to parse it read through it as such. But if that's not an option for some reason, you can look through your dictionary this way if all of your lists at each key are the same length:
for i, res in enumerate(dict[list(dict)[0]]):
ith_values = [elem[i] for elem in dict.values()]
print(ith_values)
If they're all different lengths, then you'll need to put some logic to check for that and print a blank or do some error handling for looking past the end of the list.
I'm trying to get a single key value pair using dictionary comprehension as an exercise, I have accomplished this using for loops but the best I can do using dictionary comprehension returns an entire dictionary. If I use anything other than 'inner_key01' or 'inner_key02' in the if portion of the below code I get an empty dictionary.
I would like the code to return 'inner_value22'
my_dict = {'inner_key01' :{'inner_key1': 'inner_value1', 'inner_key2': 'inner_value2'},
'inner_key02' :{'inner_key21': 'inner_value21', 'inner_key22': 'inner_value22'}
}
next_dict = {inner_key: inner_value for inner_key, inner_value in my_dict.items() for outer_key, outer_value in my_dict.items()if inner_key == 'inner_key02'}
print(next_dict)
I have a large dictionary and I am trying to remove values from keys if they start with certain values. Below is a small example of the dictionary.
a_data = {'78567908': {'26.01.17', '02.03.24', '26.01.12', '04.03.03', '01.01.13', '02.03.01', '01.01.10', '26.01.21'}, '85789070': {'26.01.02', '09.01.04', '02.05.04', '02.03.17', '02.05.01'}, '87140110': {'03.15.25', '03.15.24', '03.15.19'}, '87142218': {'26.17.13', '02.11.01', '02.03.22'}, '87006826': {'28.01.03'}}
After I read in the dictionary, I want to remove values from all keys that start with '26.' or '02.' It is possible that would leave a key with no values (an empty set).
I do have code that works:
exclude = ('26.', '02.')
f_a_data = {}
for k, v in a_data.items():
f_a_data.setdefault(k,[])
for code in v:
print (k, code, not code.startswith(exclude))
if not code.startswith(exclude):
f_a_data[k].append(code)
print('Filtered dict:')
print(f_a_data)
This returns a filtered dict:
Filtered dict:
{'78567908': ['04.03.03', '01.01.13', '01.01.10'], '85789070': ['09.01.04'], '87140110': ['03.15.25', '03.15.24', '03.15.19'], '87142218': [], '87006826': ['28.01.03']}
Question 1: Is this the best way to filter a dictionary?
Question 2: How could i modify the above snippet to return values in a set like the original dict?
Your code is quite all right in complexity terms but can be "pythonized" a little and still remain readable.
My proposal: you can rebuild a dictionary using nested comprehensions and all to test if you should include the values:
a_data = {'78567908': {'26.01.17', '02.03.24', '26.01.12', '04.03.03', '01.01.13', '02.03.01', '01.01.10', '26.01.21'}, '85789070': {'26.01.02', '09.01.04', '02.05.04', '02.03.17', '02.05.01'}, '87140110': {'03.15.25', '03.15.24', '03.15.19'}, '87142218': {'26.17.13', '02.11.01', '02.03.22'}, '87006826': {'28.01.03'}}
exclude = ('26.', '02.')
new_data = {k:{x for x in v if all(s not in x for s in exclude)} for k,v in a_data.items()}
result:
>>> new_data
{'78567908': {'01.01.10', '01.01.13', '04.03.03'},
'85789070': {'09.01.04'},
'87006826': {'28.01.03'},
'87140110': {'03.15.19', '03.15.25', '03.15.24'},
'87142218': set()}
(here using a dictionary comprehension embedding a set comprehension (since you need a set) using a generator comprehension in all)