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List output of program
print ('list', L)
['A','B']
dictionary output of program after the manipulation of program is below
print ('dictionary', d)
dictionary {'a':1,'b':2}
dictionary {'a':1,'b':2}
Expected Out
{'A':{'a':1,'b':2}, 'B': {'a':1,'b':2}}
list_out = ['A', 'B']
dict_out1 = {'a':1,,'b':2}
dict_out2 = {'a':1,,'b':2}
exp_out = {list_out[0]: dict_out1,
list_out[1]: dict_out2}
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I have a list like below
l1 = ['a10','b2','a2','c1','b4','c5']
and I want to sum of numeric and output like below
l2 = ['a12','b6','c7'].
Note: do not use in-built function
from collections import Counter
l1 = ['a10','b2','a2','c1','b4','c5']
l2 = [f'{k}{v}' for (k, v) in Counter(''.join(i[0]*int(i[1:]) for i in l1)).items()]
# ['a12', 'b6', 'c6']
Figuring this out is left as an exercise to the reader.
It also assumes the alpha part is always just the first character.
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I have this two lists:
values = ['0.7%', '11.1%', '8.7%']
key = ['NA']
which I want to zip to dict(ie. Keys and value). so that output will look like this:
{'NA': '0.7%', '11.1%', '8.7%'}
here is my current code:
values = ['0.7%', '11.1%', '8.7%']
key = ['NA']
joiner = dict(zip(key, values))
print(joiner)
currrent output:
{'NA': '0.7%'}
Completed:
values = ['0.7%', '11.1%', '8.7%']
key = ['NA']
joiner = {f"{key[0]}": values}
print(joiner)
print({key[0]:' ,'.join(values)})
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I have a list in python of the following form:
myList = ['r0x94', 'r0x21', 'r0x51']
I want to sort it based on the last number in each string entry of the list such that:
sorted_myList = ['r0x21', 'r0x51', 'r0x94']
The last number is not hex, rather it is decimal. How to do it?
>>> my_list = ['r0x94', 'r0x21', 'r0x51']
>>> sorted(my_list, key=lambda x: int(x.rpartition('x')[-1]))
['r0x21', 'r0x51', 'r0x94']
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If I have a list like:
['15', 'A:B', 'B:C', 'D:C', 'F:A']
I want to change the values like:
['1515', 'A:B15', 'B:C15', 'D:C15', 'F:A15']
(add first one to other values in a list)
Is it possible?
Assuming ["15", "A:B", "B:C", "D:C", "F:A"] is your input list, a simple list comprehension would do the job:
>>> l = ["15", "A:B", "B:C", "D:C", "F:A"]
>>> value_to_add = l[0]
>>> [item + value_to_add for item in l]
['1515', 'A:B15', 'B:C15', 'D:C15', 'F:A15']
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I'm trying to convert a dictionary
clues={#:A,+:B,6:C}
into
clues1=[#,+,6]
clues2=[A,B,C]
when I use clues.values it doesn't let me iterate through it and any other method I have used has given an error message
Do you want to set keys to clues1 and values to clues2?
clues1 = list(clues.keys())
clues2 = list(clues.values())
If you want the lists to be in alphabetical order, but still have the indicies match, you can do:
clues={'#':'A','+':'B','6':'C'}
cluesKeys = list(clues.iterkeys())
cluesValues = list(clues.itervalues())
If you explicitly want an iteration over the "clues" dictionary:
clues={'#':'A','+':'B','6':'C'}
clues1=list()
clues2=list()
for keys, values in clues.items():
clues1.append(keys)
clues2.append(values)
print clues1
print clues2
['#', '+', '6']
['A', 'B', 'C']