I have a dictionary that looks like:
{u'message': u'Approved', u'reference': u'A71E7A739E24', u'success': True}
I would like to retrieve the key-value pair for reference, i.e. { 'reference' : 'A71E7A739E24' }.
I'm trying to do this using iteritems which does return k, v pairs, and then I'm adding them to a new dictionary. But then, the resulting value is unicode rather than str for some reason and I'm not sure if this is the most straightforward way to do it:
ref = {}
for k, v in charge.iteritems():
if k == 'reference':
ref['reference'] = v
print ref
{'reference': u'A71E7A739E24'}
Is there a built-in way to do this more easily? Or, at least, to avoid using iteritems and simply return:
{ 'reference' : 'A71E7A739E24' }
The trouble with using iteritems is that you increase lookup time to O(n) where n is dictionary size, because you are no longer using a hash table
If you only need to get one key-value pair, it's as simple as
ref = { key: d[key] }
If there may be multiple pairs that are selected by some condition,
either use dict from iterable constructor (the 2nd version is better if your condition depends on values, too):
ref = dict(k,d[k] for k in charge if <condition>)
ref = dict(k,v for k,v in charge.iteritems() if <condition>)
or (since 2.7) a dict comprehension (which is syntactic sugar for the above):
ref = {k,d[k] for k in charge if <condition>}
<same as above>
I dont understand the question:
is this what you are trying to do:
ref={'reference',charge["reference"]}
Related
I want to replace items in a list based on another list as reference.
Take this example lists stored inside a dictionary:
dict1 = {
"artist1": ["dance pop","pop","funky pop"],
"artist2": ["chill house","electro house"],
"artist3": ["dark techno","electro techno"]
}
Then, I have this list as reference:
wish_list = ["house","pop","techno"]
My result should look like this:
dict1 = {
"artist1": ["pop"],
"artist2": ["house"],
"artist3": ["techno"]
}
I want to check if any of the list items inside "wishlist" is inside one of the values of the dict1. I tried around with regex, any.
This was an approach with just 1 list instead of a dictionary of multiple lists:
check = any(item in artist for item in wish_list)
if check == True:
artist_genres.clear()
artist_genres.append()
I am just beginning with Python on my own and am playing around with the SpotifyAPI to clean up my favorite songs into playlists. Thank you very much for your help!
The idea is like this,
dict1 = { "artist1" : ["dance pop","pop","funky pop"],
"artist2" : ["house","electro house"],
"artist3" : ["techno","electro techno"] }
wish_list = ["house","pop","techno"]
dict2={}
for key,value in dict1.items():
for i in wish_list:
if i in value:
dict2[key]=i
break
print(dict2)
A regex is not needed, you can get away by simply iterating over the list:
wish_list = ["house","pop","techno"]
dict1 = {
"artist1": ["dance pop","pop","funky pop"],
"artist2": ["chill house","electro house"],
"artist3": ["dark techno","electro techno"]
}
dict1 = {
# The key is reused as-is, no need to change it.
# The new value is the wishlist, filtered based on its presence in the current value
key: [genre for genre in wish_list if any(genre in item for item in value)]
for key, value in dict1.items() # this method returns a tuple (key, value) for each entry in the dictionary
}
This implementation relies a lot on list comprehensions (and also dictionary comprehensions), you might want to check it if it's new to you.
Is there a way to sort all the keys, sub-keys, sub-sub-keys, etc. of a python dictionary at once?
Let's suppose I have the dictionary
dict_1 = {
"key9":"value9",
"key5":"value5",
"key3":{
"key3_1":"value3_1",
"key3_3":"value3_3",
}
"key4":"value4",
"key2":"value2",
"key8":{
"key8_1":"value8_1",
"key8_5":[
"value8_5_3",
"value8_5_1",
]
"key8_2":"value8_2",
}
"key4":"value4",
"key1":"value1",
}
and I want it sorted as
dict_1 = {
"key1":"value1",
"key2":"value2",
"key3":{
"key3_1":"value3_1",
"key3_3":"value3_3",
}
"key4":"value4",
"key5":"value5",
"key8":{
"key8_1":"value8_1",
"key8_2":"value8_2",
"key8_5":[
"value8_5_1",
"value8_5_3",
]
}
"key9":"value9",
}
Is there a method to do it?
Please note:
potentially my dict_1 could have several levels of subkeys (nested
dictionaries) or subvalues (nested lists).
I am using Python 2.7.17, and I cannot update it. But order is not
preserved in dictionaries of Python versions previous to 3.7, so I bet
the sorting has to be done by using the OrderedDict.
First, it is important to know that dictionaries are not ordered. So, if you want to order a dict, you need to go with collections.OrderedDict (which exists since Python 2.7).
And then, this is a use case for a recursive function:
from collections import OrderedDict
def order_dict(d):
ordered_dict = OrderedDict()
for key in sorted(d.keys()):
val = d[key]
if isinstance(val, dict):
val = order_dict(val)
ordered_dict[key] = val
return ordered_dict
I am struggling to figure out an assignment.
The problem set up is:
I have a list containing ratios ( unique_ratio = ['0.05', '0.98', '1.45']
I have a dictionary containing k:v as ratio:count the number of times ratio has appeared in a previous variable ( dict = {'0.05':'5', '0.32':'72', '0.98': '21'}
I want to iterate over the dictionary and extract the k:v for the ratios which appear in the unique_ratio list. I want to store these k:v's in a new dictionary (frequencies = {})
I am running pytho 3.7
I have tried iterating over the dictionary using for loop but am never able to extract the k:v pair.
I am unsure whether I should test for i in unique_ratios or i in dict
for i in dict.values():
frequencies = { k:v for k,v in comp_dict_count.items() if 'i' in
unique_ratios }
print(frequencies)
Everything I have tried has led to syntax errors. The above code leads to empty frequencies dictionary.
You need a single dictionary comprehension for this. Also for a better formormance you could check membership using sets, reducing the lookup complexity to O(1):
unique_ratio = set(['0.05', '0.98', '1.45'])
d = {'0.05':'5', '0.32':'72', '0.98': '21'}
{k:v for k,v in d.items() if k in unique_ratio}
# {'0.05': '5', '0.98': '21'}
Let's say I have
my_dict = {
("airport", "London"): "Heathrow",
("airport", "Tokyo"): "Narita",
("hipsters", "London"): "Soho"
}
What is an efficient (no scanning of all keys), yet elegant way to get all airports out of this dictionary, i.e. expected output ["Heathrow", "Narita"]. In databases that can index by tuples, it's usually possible to do something like
airports = my_dict.get(("airport",*))
(but usually only with the 'stars' sitting at the rightmost places in the tuple since the index usually is only stored in one order).
Since I imagine Python to index dictionary with tuple keys in a similar way (using the keys's inherent order), I imagine there might be a method I could use to slice the index this way?
Edit1: Added expected output
Edit2: Removed last phrase. Added '(no scanning of all keys)' to the conditions to make it clearer.
The way your data is currently organized doesn't allow efficient lookup - essentially you have to scan all the keys.
Dictionaries are hash tables behind the scenes, and the only way to access a value is to get the hash of the key - and for that, you need the whole key.
Use a nested hierarchy like this, so you can do a direct O(1) lookup:
my_dict = {
"airport": {
"London": "Heathrow",
"Tokyo": "Narita",
},
"hipsters": {
"London": "Soho"
}
}
Check "airport" is present in the every key in the dictionary.
Demo:
>>> [value for key, value in my_dict.items() if "airport" in key]
['Narita', 'Heathrow']
>>>
Yes, Nested dictionary will be better option.
>>> my_dict = {
... "airport": {
... "London": "Heathrow",
... "Tokyo": "Narita",
... },
... "hipsters": {
... "London": "Soho"
... }
... }
>>>
>>> if "airport" in my_dict:
... result = my_dict["airport"].values()
... else:
... result = []
...
>>> print result
['Heathrow', 'Narita']
>>>
What I'd like to avoid, if possible, is to go through all dictionary keys and filter them down.
Why? Why do you think Python is doing the equivalent of a DB full table scan? Filtering a dictionary does not mean sequential scanning it.
Python:
[value for key, value in my_dict.items() if key[0] == "airport"]
Output:
['Narita', 'Heathrow']
I am facing some trouble with trying to add more key:value pairs to a dictionary object that is itself nested within another dictionary object. Also, the usual way of doing dict[key] = value to assign additional key:value pairs to the dictionary is not suitable for my case here (I'll explain why later below), and thus this makes my objective a lot more challenging to achieve.
I'll illustrate what I'm trying to achieve with some statements from my source code.
First, I have a dictionary object that contains nesting:
environment = { 'v' :
{
'SDC_PERIOD':'{period}s'.format(period = self.period),
'FAMILY':'{legup_family}s'.format(legup_family = self.legup_family),
'DEVICE_FAMILY':'"{fpga_family}s"'.format(fpga_family = self.fpga_family)
}
}
and then following this line, I will do an if test that, if passed, will require me to add this other dictionary:
environment_add = { 'v' : {'LM_LICENSE_FILE' : '1800#adsc-linux'} ,
'l' : 'quartus_full' }
to ultimately form this complete dictionary:
environment = { 'v' :
{
'SDC_PERIOD':'{period}s'.format(period = self.period),
'FAMILY':'{legup_family}s'.format(legup_family = self.legup_family),
'DEVICE_FAMILY':'"{fpga_family}s"'.format(fpga_family = self.fpga_family),
'LM_LICENSE_FILE' : '1800#adsc-linux'
} ,
'l' : 'quartus_full'
}
As you can see, if I were to try and assign a new key:value pair using the dict[key] = value syntax, it would not work for me because it would end up either creating an new key:value pair for me, or overwrite the existing dictionary object and the key:value pairs that are nested under the 'v' key.
So far, in order to accomplish the creation of the dictionary, I've been using the following:
environment = """{ v: {'SDC_PERIOD':'%(period)s','FAMILY':'%(legup_family)s','DEVICE_FAMILY':'"%(fpga_family)s"'}}""" % self
if self.require_license: # this is the if statement test that I was referring to earlier
environment = environment.replace('}', '')
environment += """ ,'LM_LICENSE_FILE':'1800#adsc-linux'}, 'l': 'quartus_full'}"""
and then obtaining the dictionary object later with:
import ast
env_dict = ast.literal_eval(environment)
which gives effectively converts the environment string into a dictionary object stored under a new variable name of env_dict.
My teammates think that this is much too overkill, especially since the environment or env_dict object will be parsed in 2 separate modules later on. In the first module, the key-value pairs will be broken up and reconstructed to form strings that look like '-v SDC_PERIOD=2500s, LM_LICENSE_FILE=1800#adsc-linux' , while in the second module, the dictionary nested under the 'v' key (of the environment/env_dict dictionary object) will be extracted out and directly fed as an argument to a function that accepts a mapping object.
So as you can see, there is quite a lot of precise parsing required to do the job, and although my method fulfills the objective, it is not accepted by my team and they think that there must be a better way to do this directly from environment being a dictionary object and not a string object.
Thank you very much for studying my detailed post, and I will greatly appreciate any help or suggestions to move forward on this!
for k,v in environment_add.iteritems(): # .items() in Python 3
if k in environment:
environment[k].update(v)
else:
environment[k] = v
That is, for each item to add, check if it exists, and update it if so, or simply create it. This assumes the items being added, if they exist, will be dicts (you can't update a string like quartus_full).
Why not just use update
In [4]: dict_ = {"a": {"b": 2, "c": 3}}
In [5]: dict_["a"].update(d=4)
In [6]: dict_
Out[6]: {'a': {'b': 2, 'c': 3, 'd': 4}}