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I am working with dictionaries and want to solve this custom problem wherein I want values in my dictionary to be in the same position as in my_list. In case, we don't have the value in the dictionary it should place with '-'
All the values in the dictionary are available in my_list.
my_list = ['a', 'b', 'c', 'd']
d = {'x':['a', 'b', 'd'], 'y':['a', 'c'], 'z': ['d', 'b']}
expected output:
{'x':['a', 'b', '-', 'd'], 'y':['a', '-', 'c', '-'], 'z':['-', 'b', '-', 'd']}
You can iterate over my_list and check each value exists in list that exist in d. We can change the dict in-place like below:
my_list = ['a', 'b', 'c', 'd']
d = {'x':['a', 'b', 'd'], 'y':['a', 'c'], 'z': ['d', 'b']}
for key, lst in d.items():
d[key] = [val if val in lst else '-' for val in my_list]
print(d)
# {'x': ['a', 'b', '-', 'd'], 'y': ['a', '-', 'c', '-'], 'z': ['-', 'b', '-', 'd']}
from collections import OrderedDict
my_list=['a','b','c','d']
d={'x':['a','b','d'],'y':['a','c'],'z':['d','b']}
z=list(d)
new=OrderedDict.fromkeys(z,'-')
for i in d.keys():
dict = OrderedDict.fromkeys(my_list,'-')
for j in d[i]:
dict[j]=j
new[i]=list(dict.values())
print(new)
I have a list:
['A', 'B', 'C', ['D', ['E', 'F'], 'G'], 'H']
and I want to turn this into:
[['E', 'F'], ['D', 'G'], ['A', 'B', 'C', 'H']]
So basically I want the sublist on the deepest level of the list to come first in the new list and then counting down the level the remaining sublists.
This should work with any nested list.
If there are two sublists on the same level, then it doesn't really matter which one comes first.
['A', 'B', 'C', ['D', ['E', 'F'], 'G'], ['H', 'I', 'J']]
[['E', 'F'], ['D', 'G'], ['H', 'I', 'J'], ['A', 'B', 'C', 'H']] #this is fine
[['E', 'F'], ['H', 'I', 'J'], ['D', 'G'], ['A', 'B', 'C', 'H']] #this too
I thought of first using a function to determine on what level the deepest sublist is, but then again I don't know how to access items in a list based on their level or if that's even possible.
Been tinkering around this for far too long now and I think my head just gave up, hope someone can assist me with this problem!
You can use a recursive generator function:
def sort_depth(d, c = 0):
r = {0:[], 1:[]}
for i in d:
r[not isinstance(i, list)].append(i)
yield from [i for j in r[0] for i in sort_depth(j, c+1)]
yield (c, r[1])
def result(d):
return [b for _, b in sorted(sort_depth(d), key=lambda x:x[0], reverse=True) if b]
print(result(['A', 'B', 'C', ['D', ['E', 'F'], 'G'], 'H']))
print(result(['A', 'B', 'C', ['D', ['E', 'F'], 'G'], ['H', 'I', 'J']]))
print(result([[1, [2]], [3, [4]]]))
Output:
[['E', 'F'], ['D', 'G'], ['A', 'B', 'C', 'H']]
[['E', 'F'], ['D', 'G'], ['H', 'I', 'J'], ['A', 'B', 'C']]
[[2], [4], [1], [3]]
Here is a relatively straight-forward solution:
def sort_depth(d):
def dlists(obj, dep=0):
for x in filter(list.__instancecheck__, obj):
yield from dlists(x, dep-1)
yield [x for x in obj if not isinstance(x, list)], dep
return [x for x, y in sorted(dlists(d), key=lambda p: p[1])]
>>> [*sort_depth([[1, [2]], [3, [4]]])]
[[2], [4], [1], [3], []]
>>> [*sort_depth(['A', 'B', 'C', ['D', ['E', 'F'], 'G'], 'H'])]
[['E', 'F'], ['D', 'G'], ['A', 'B', 'C', 'H']]
The approach:
Collect all the sublists and annotate them with their (negative) nesting level, e.g. (['E', 'F'], -2)
Sort them by their nesting level
Extract the lists back from the sorted data
Say I have this input data
my_input_list = [[A],[A,B,C],[D],[D,E,F],[A,B,C,D,E,F],[A,C,E]]
items_that_appear_twice = [A,B,C]
items_that_appear_four = [D,E,F]
And I want to create an expansion such that some elements are only allowed to appear twice.
my_output_list = [
[A],[A],
[A,B,C],[A,B,C],
[D],[D],[D],[D],
[D,E,F],[D,E,F],[D,E,F],[D,E,F],
[A,B,C,D,E,F],[A,B,C,D,E,F],[D,E,F],[D,E,F],
[A,C,E],[A,C,E],[E],[E]]
I tired a few ideas and didn't find a really neat solution, like building lists of four and list.remove() from them which generated two empty lists.
For example list removal techniques on my_input_list[0]*4 gives [A],[A],[],[] (two empty lists) when I want [A],[A] .
I have a working version: See Pyfiddle.
my_input_list = [['A'],['A','B','C'],['D'],['D','E','F'],['A','B','C','D','E','F'],['A','C','E']]
items_that_appear_twice = ['A','B','C']
items_that_appear_four = ['D','E','F']
my_output_list = []
for my_input in my_input_list:
items_allowed_to_appear_twice = list(filter(
lambda value: (value in items_that_appear_twice
or value in items_that_appear_four),
my_input))
items_allowed_to_appear_four = list(filter(
lambda value: value in items_that_appear_four,
my_input))
my_output_list += 2*[items_allowed_to_appear_twice]
if len(items_allowed_to_appear_four):
my_output_list += 2*[items_allowed_to_appear_four]
print(my_output_list)
my_input_list = [['A'],
['A', 'B', 'C'],
['D'],
['D', 'E', 'F'],
['A', 'B', 'C', 'D', 'E', 'F'],
['A', 'C', 'E']]
items_that_appear_twice = ['A', 'B', 'C']
items_that_appear_four = ['D', 'E', 'F']
my_output_list = []
for sub in my_input_list:
my_output_list.append(sub)
my_output_list.append(sub)
sub = [x for x in sub if x in items_that_appear_four]
if sub:
my_output_list.append(sub)
my_output_list.append(sub)
assert my_output_list == [
['A'], ['A'],
['A', 'B', 'C'], ['A', 'B', 'C'],
['D'], ['D'], ['D'], ['D'],
['D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F'],
['A', 'B', 'C', 'D', 'E', 'F'], ['A', 'B', 'C', 'D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F'],
['A', 'C', 'E'], ['A', 'C', 'E'], ['E'], ['E']]
Below is my solution, sort the sub list first, then append to the result according different situation.
my_input_list = [['A'],['A','B','C'],['D'],['D','E','F'],['A','B','C','D','E','F'],['A','C','E']]
items_that_appear_twice = ['A','B','C']
items_that_appear_four = ['D','E','F']
result = []
for sublist in my_input_list:
appearence = {1:[], 2:[], 4:[]}
for item in sublist:
appearence[2].append(item) if item in items_that_appear_twice else (appearence[4].append(item) if item in items_that_appear_four else appearence[1].append(item))
if len(appearence[2]) > 0:
result.append([appearence[2] + appearence[4]] * 2 + ([appearence[4]] * 2 if appearence[4] and len(appearence[4]) > 0 else []))
else:
result.append([appearence[4]] * 4)
for item in result:
print(item)
output:
[['A'], ['A']]
[['A', 'B', 'C'], ['A', 'B', 'C']]
[['D'], ['D'], ['D'], ['D']]
[['D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F']]
[['A', 'B', 'C', 'D', 'E', 'F'], ['A', 'B', 'C', 'D', 'E', 'F'], ['D', 'E', 'F'], ['D', 'E', 'F']]
[['A', 'C', 'E'], ['A', 'C', 'E'], ['E'], ['E']]
[Finished in 0.178s]
Say you have a dictionary describing item dependencies, along the lines of:
deps = {
'A': ['B', 'C', 'D'],
'B': ['C', 'E'],
'C': ['D', 'F'],
'D': ['C', 'G'],
'E': ['A'],
'H': ['N'],
}
meaning that item 'A' depends on items 'B', 'C', and 'D', etc. Obviously, this could be of arbitrary complexity.
How do you write a function get_all_deps(item) that gives you a list of all the dependencies of item, without duplicates and without item. E.g.:
> get_all_deps('H')
['N']
> get_all_deps('A')
['B', 'C', 'D', 'E', 'F', 'G']
> get_all_deps('E')
['A', 'B', 'C', 'D', 'F', 'G']
I'm looking for concise code - ideally a single recursive function. Performance is not terribly important for my use case - we're talking about fairly small dependency graphs (e.g. a few dozen items)
you can use a stack/todo list to avoid recursive implementation:
deps = {
'A': ['B', 'C', 'D'],
'B': ['C', 'E'],
'C': ['D', 'F'],
'D': ['C', 'G'],
'E': ['A'],
'H': ['N'],
}
def get_all_deps(item):
todo = set(deps[item])
rval = set()
while todo:
subitem = todo.pop()
if subitem != item: # don't add start item to the list
rval.add(subitem)
to_add = set(deps.get(subitem,[]))
todo.update(to_add.difference(rval))
return sorted(rval)
print(get_all_deps('A'))
print(get_all_deps('E'))
print(get_all_deps('H'))
result:
['B', 'C', 'D', 'E', 'F', 'G']
['A', 'B', 'C', 'D', 'F', 'G']
['N']
todo set contains the elements to be processed.
Pop one element and put it in return value list
Loop until no more elements (okay there's a loop in here)
add only the elements to process if they're not already in the return value.
return sorted list
The set difference avoids the problem with cyclic dependencies, and the "max recursion depth" is avoided. Only limit is system memory.
Currently working on a 2D transposition cipher in Python. So I have a list that contains an encoded message, like below:
['BF', 'AF', 'AF', 'DA', 'CD', 'DD', 'BC', 'EF', 'DA', 'AA', 'EF', 'BF']
The next step is taking that list, splitting it up and putting it into a new matrix with regards to a keyword that the user enters. Which I have below:
Enter the keyword for final encryption: hide
H I D E
['B', 'F', 'A', 'F']
['A', 'F', 'D', 'A']
['C', 'D', 'D', 'D']
['B', 'C', 'E', 'F']
['D', 'A', 'A', 'A']
['E', 'F', 'B', 'F']
What I would like to do next and haven't done is take each of the columns above and print them in alphabetical order, therefore getting another cipher text, like below:
D E H I
['A', 'F', 'B', 'F']
['D', 'A', 'A', 'F']
['D', 'D', 'C', 'D']
['E', 'F', 'B', 'C']
['A', 'A', 'D', 'A']
['B', 'F', 'E', 'F']
Here's my code:
def encodeFinalCipher():
matrix2 = []
# Convert keyword to upper case
keywordKey = list(keyword.upper())
# Convert firstEncryption to a string
firstEncryptionString = ''.join(str(x) for x in firstEncryption)
# Print the first table that will show the firstEncryption and the keyword above it
keywordList = list(firstEncryptionString)
for x in range(0,len(keywordList),len(keyword)):
matrix2.append(list(keywordList[x:x+len(keyword)]))
# Print the matrix to the screen
print (' %s' % ' '.join(map(str, keywordKey)))
for letters in matrix2:
print (letters)
return finalEncryption
I have traversed the 2D matrix and got all the column entries like below:
b = [[matrix2[i][j] for i in range(len(matrix2))] for j in range(len(matrix2[0]))]
for index, item in enumerate (b):
print("\n",index, item)
OUTPUT:------
0 ['B', 'A', 'C', 'B', 'D', 'E']
1 ['F', 'F', 'D', 'C', 'A', 'F']
2 ['A', 'D', 'D', 'E', 'A', 'B']
3 ['F', 'A', 'D', 'F', 'A', 'F']
How would I append each letter of the keywordKey (e.g. 'H' 'I' 'D' 'E') to the list where the numbers 0,1,2,3 are?
Or probably a more efficient solution. How would I put the letters into the keywordKey columns when creating the matrix? Would a dictionary help here? Then I could sort the dictionary and print the final cipher.
Many thanks
You can do something like this:
>>> from operator import itemgetter
>>> from pprint import pprint
>>> lst = [['B', 'F', 'A', 'F'],
['A', 'F', 'D', 'A'],
['C', 'D', 'D', 'D'],
['B', 'C', 'E', 'F'],
['D', 'A', 'A', 'A'],
['E', 'F', 'B', 'F']]
>>> key = 'HIDE'
Sort xrange(len(key)) or range(len(key)) using the corresponding values from key and then you will have a list of indices:
>>> indices = sorted(xrange(len(key)), key=key.__getitem__)
>>> indices
[2, 3, 0, 1]
Now all we need to do is loop over the list and apply these indices to each item using operator.itemgetter and get the corresponding items:
>>> pprint([list(itemgetter(*indices)(x)) for x in lst])
[['A', 'F', 'B', 'F'],
['D', 'A', 'A', 'F'],
['D', 'D', 'C', 'D'],
['E', 'F', 'B', 'C'],
['A', 'A', 'D', 'A'],
['B', 'F', 'E', 'F']]
#or simply
>>> pprint([[x[i] for i in indices] for x in lst])
[['A', 'F', 'B', 'F'],
['D', 'A', 'A', 'F'],
['D', 'D', 'C', 'D'],
['E', 'F', 'B', 'C'],
['A', 'A', 'D', 'A'],
['B', 'F', 'E', 'F']]