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How to neatly print dictionaries with dictionaries inside


How to neatly print dictionaries with dictionaries inside

By : harsha
Date : November 21 2020, 11:01 PM
Hope this helps iteritems only produces two pairs - key and value - at a time, so you can't unpack into three items in that for loop.
Instead, you want to do:
code :
for name in sick:
     print name, ": ", sick[name]['Reason']
     print "Sick chit expires: ", sick[name]['Expiry']


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How create dictionaries inside dictionaries, inside dictionaries... 'n' times

How create dictionaries inside dictionaries, inside dictionaries... 'n' times


By : sgraewe
Date : March 29 2020, 07:55 AM
may help you . I need a way to create n-dimensional dictionaries. , Execute this doint_it() over and over:
code :
def doing_it(dic):
    new_mod = {}
    for i in sorted(dic.keys()):
        if dic[i] == 1:
            new_mod[i] = dic1
        elif dic[i] == 2:
            new_mod[i] = dic2
        else:
            new_mod[i] = doing_it(dic[i])
    return new_mod
Parse Json nested dictionaries with Python (not dictionaries inside lists)

Parse Json nested dictionaries with Python (not dictionaries inside lists)


By : chaoran li
Date : March 29 2020, 07:55 AM
this one helps. I have these horrible nested JSON dictionaries: , you probably want try: except like
code :
try:
    assert x in data.keys() for x in ["x","y"]
    ...
    return data["x"]["y"]
except:
    return None
Python 3, add two dictionaries and adding item inside the list associated with a key both dictionaries have

Python 3, add two dictionaries and adding item inside the list associated with a key both dictionaries have


By : Gavin Olsen
Date : March 29 2020, 07:55 AM
I hope this helps you . If you are happy using a 3rd party library, you can do this easily with numpy.
I have intentionally kept the output in numeric form. It is good practice to store numeric data in numeric types.
code :
import numpy as np

d1= {"name_1":["0.0","0.0","0.0","4.1","2.2"],
     "name_2":["8.4","7.3","7.2","4.0","2.0"],
     "name_3":["7.0","7.0","5.0","6.0","3.0"]}

d2 = {"name_1":["8.0","8.0","8.0","0.0","3.0"]}

res = {k: np.array(d1.get(k, [0]*5), dtype=float) + \
          np.array(d2.get(k, [0]*5), dtype=float) \
       for k in d1.keys() | d2.keys()}

# {'name_1': array([ 8. ,  8. ,  8. ,  4.1,  5.2]),
#  'name_2': array([ 8.4,  7.3,  7.2,  4. ,  2. ]),
#  'name_3': array([ 7.,  7.,  5.,  6.,  3.])}
Match values of two dictionaries and update list inside one of the dictionaries with matching key value pair

Match values of two dictionaries and update list inside one of the dictionaries with matching key value pair


By : Sumit
Date : March 29 2020, 07:55 AM
will be helpful for those in need Hello have tried to figure this out for some time without much luck so any help is greatly appreciated. , The result is:
code :
result = [{'link': 'www.home.com', 'title': [{'one': 1, 'two': 2, 'three': 3}]}, {'link': 'www.away.com', 'title': [{'two': 2, 'three': 3}]}]
from copy import deepcopy

aa = [{'link': 'www.home.com', 'title': ['one', 'two', 'three']}, {'link': 'www.away.com', 'title':['two', 'three']}]
bb = [{'id': 1, 'title' :'one'},{'id': 2, 'title': 'two'}, {'id': 3, 'title': 'three'}]

titleids = {}
for b in bb:
    titleids[b['title']] = b['id']

result = deepcopy(aa)
for a in result:
    a['title'] = [{title:titleids[title] for title in a['title']}]

print(result)
Dictionary of dictionaries: print dictionaries that share at least two common keys

Dictionary of dictionaries: print dictionaries that share at least two common keys


By : phani kumar vadlaman
Date : March 29 2020, 07:55 AM
I wish this helpful for you I hope I understood correctly what you want. The approach is clumsy and I fear it is highly inefficient.
I added a dictionary g6 to d in order to produce a more interesting output:
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