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Access dict keys and list elements by same index to loop over and assign values


Access dict keys and list elements by same index to loop over and assign values

By : user3042614
Date : November 28 2020, 12:01 PM
will help you To start with, unless you are on python 3.6=>, dict will not guarantee you the order. In order words, hdA1c will not always be the first key.
If you were to maintain the order, use collections.OrderedDict.
code :
for i, (k, v) in enumerate(dataframesdict.items()):
    v['unit'] = unit_list[i]

print(dataframesdict['hbA1c'])
                  Date  Value  person_id unit
0  2009-01-05 00:00:00    9.6          1    %
1  2009-01-05 00:00:00    9.6          2    %
2  2013-01-31 00:00:00   10.0          3    %


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What's a fast (non-loop) way to apply a dict to a ndarray (meaning use elements as keys and replace with values)

What's a fast (non-loop) way to apply a dict to a ndarray (meaning use elements as keys and replace with values)


By : ogunkarakus
Date : March 29 2020, 07:55 AM
it fixes the issue At the moment I'm looping through like , transform your dictionary in array and use np.take:
code :
Na=1000 #array
Nd = 10**4 #dict
data=randint(0,Nd,(Na,Na))
dic=dict(zip(range(Nd),randint(0,Nd,Nd)))
dicarray=np.array(list(dic.values()))
In [3]: %timeit np.array([[dic[x] for x in line] for line in data]) 
1 loops, best of 3: 2.27 s per loop

In [4]: %timeit dicarray.take(data)
10 loops, best of 3: 24.4 ms per loop
Python assignment quirk w/ list index assign, dict index assign, and dict.get

Python assignment quirk w/ list index assign, dict index assign, and dict.get


By : Viah Reyes
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Ruby and Python are different languages, and make different choices. In Python assignments are statements and evaluates multiple assignment targets from left to right. Ruby made other choices; assignments are expressions and as a result are evaluated in the opposite order.
So in Ruby this happens:
code :
if a = 42:
if a == 42:
Dictionary/list comprehension to assign new dict values from lists to predefined keys

Dictionary/list comprehension to assign new dict values from lists to predefined keys


By : PtrSimek
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I have 2 lists containing strings (they could also be changed to sets instead, if that makes a difference here): , You can use zip + list/dict comprehensions:
code :
names = ['abc' , 'xyz' , 'foo']
ids = ['1234' , '5678' , '9101']

res = [{"name": n, "id": i} for n, i in zip(names, ids)]
Insert to dict multiple keys from list with for loop in one line - dict.setdefault

Insert to dict multiple keys from list with for loop in one line - dict.setdefault


By : Mohit Engg
Date : March 29 2020, 07:55 AM
I hope this helps you . I tried to do this: , You can squish a for loop with a one-line body into one line:
code :
ip_dict = dict()
ip_list = ["10.0.0.1", "10.0.0.2"]
for i in ip_list: ip_dict.setdefault(i, 0)
ip_list = ["10.0.0.1", "10.0.0.2"]
ip_dict = dict.fromkeys(ip_list, 0)
How to assign the values for dict(keys and values) into a new list? and also how to print the keys and values in the str

How to assign the values for dict(keys and values) into a new list? and also how to print the keys and values in the str


By : ram
Date : March 29 2020, 07:55 AM
I wish this helpful for you One possible solution is to filter out non-dictionaries using list comprehension, then convert the dictionaries into key-value tuples, and separate the keys and values with zip:
code :
k,v = zip(*[list(x.items())[0] for x in a if isinstance(x, dict)])
print(k,v)
#('1', '9', '4', '3', '0') ('0', '2', '3', '5', '7')
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