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Invert keys and values of the original dictionary
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Closed 8 years ago.
I am looking to tranpose a dictionary on python and after looking around i was not able to ifnd a solution for this. Does anybody know how could i reverse a dictionary like the following as input:
graph = {'A': ['B', 'C'],
'B': ['C', 'D'],
'C': ['D'],
'D': ['C'],
'E': ['F'],
'F': ['C']}
so that i get something like:
newgraph = {'A': [''],
'B': ['A'],
'C': ['A', 'B', 'D','F'],
'D': ['B', 'C'],
'E': [''],
'F': ['E']}
Use defaultdict:
newgraph = defaultdict(list)
for x, adj in graph.items():
for y in adj:
newgraph[y].append(x)
While it doesn't seem to make any sense to have the empty string '' in the empty lists, it's certainly possible:
for x in newgraph:
newgraph[x] = newgraph[x] or ['']
Use defaultdict:
>>> from collections import defaultdict
>>> graph = {'A': ['B', 'C'],
... 'B': ['C', 'D'],
... 'C': ['D'],
... 'D': ['C'],
... 'E': ['F'],
... 'F': ['C']}
>>> new_graph = defaultdict(list)
>>> for ele in graph.keys():
... new_graph[ele] = []
...
>>> for k, v in graph.items():
... for ele in v:
... new_graph[ele].append(k)
...
>>> pprint(new_graph)
{'A': [],
'B': ['A'],
'C': ['A', 'B', 'D', 'F'],
'D': ['B', 'C'],
'E': [],
'F': ['E']}
It's also possible without defaultdict.
Here I've left the empty keys in the new dict with the value None.
graph = {'A': ['B', 'C'],
'B': ['C', 'D'],
'C': ['D'],
'D': ['C'],
'E': ['F'],
'F': ['C']}
g = dict.fromkeys(graph.keys())
for k, v in graph.iteritems():
for x in v:
if g[x]: g[x] += [k]
else: g[x] = [k]
for k in sorted(graph.keys()):
print k, ':', g[k]
Output:
A : None
B : ['A']
C : ['A', 'B', 'D', 'F']
D : ['C', 'B']
E : None
F : ['E']
Related
def recursion(input_type):
print('input_type ',input_type)
if isinstance(input_type, dict):
num = 0
for k,v in input_type.items():
if isinstance(v, dict):
print('from recursion')
recursion(v)
elif isinstance(v, list):
for j in v:
if isinstance(j, dict):
print('from recursion level 2')
recursion(j)
else:
temp_dict = {k:v}
print('type: ',type(temp_dict), k, v)
print('num',num)
num = num+1
for i in list_:
recursion(i)
How to get the interim results from the recursion.
consider the input as shown below:
input: [{'a':a, 'b':b, 'c':[{'d':d, 'e':e}]}]
Updated input: [ {'a':a, 'b':b, 'c': { 'd':d, 'e': [ {'f':f, 'g':g}, {'f':f1, 'g':g1} ] } } ]
desired output: [{'a':a, 'b':b, 'd':d, 'f':f, 'g':g, 'f_new':f1, 'g_new':g1}]
If the key is duplicate then it should update such as 'f' to 'f_new' or something like that
Thank you in advance!!
You can iterate over the dict items and if an item is a list, recursively flatten the dicts within it:
def f(o):
return {a: b for k, v in o.items() for a, b in ((i for d in (v if isinstance(v, list)
else (v,)) for i in f(d).items()) if isinstance(v, (list, dict)) else ((k, v),))}
so that given:
lst = [{'a': 'a', 'b': 'b', 'c': [{'d': 'd', 'e': 'e'}, {'f': [{'g': 'g'}]}]}]
[f(d) for d in lst] would return:
[{'a': 'a', 'b': 'b', 'd': 'd', 'e': 'e', 'g': 'g'}]
and that given:
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': {'f': 'f', 'g': 'g'}}}]
[f(d) for d in lst] would return:
[{'a': 'a', 'b': 'b', 'd': 'd', 'f': 'f', 'g': 'g'}]
To avoid collisions in merged keys, append _new to a duplicating key until it is found not pre-existing, in which case you cannot use comprehension:
def f(o):
output = {}
for k, v in o.items():
for a, b in ((i for d in (v if isinstance(v, list) else (v,)) for i in f(d).items())
if isinstance(v, (list, dict)) else ((k, v),)):
while a in output:
a += '_new'
output[a] = b
return output
so that given:
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': [{'f': 'f', 'g': 'g'}, {'f': 'f1', 'g': 'g1'}]}}]
[f(d) for d in lst] would return:
[{'a': 'a', 'b': 'b', 'd': 'd', 'f': 'f', 'g': 'g', 'f_new': 'f1', 'g_new': 'g1'}]
Demo: https://repl.it/#blhsing/NonstopSeveralActionscript
I need to store records arranged in ascending order, with log(n) access time. I come from C++ background and if I had to use C++, I would have gone for std::map which implements red-black tree internally. This guaranties the records to always be stored in ascending order of the keys, and also guaranties log(n) access time. But what's the best way to do this in Python3.5?
One way to solve this problem will be to use the bintrees library, but is there a dedicated library for storing sorted records?
You can use sortedContainers which can allow you maintain an always sorted data-structures (list, dict, dictWithKeys, set).
You can install using
pip install sortedcontainers
Here is a quick example
import sortedcontainers
g = {'B': ['A', 'C'],
'C': ['D'],
'A': ['B', 'C'],
'D': [],
}
l = sortedcontainers.SortedDict(g)
>>> l
SortedDict(None, 1000, {'A': ['B', 'C'], 'B': ['A', 'C'], 'C': ['D'], 'D': []})
>>> l['G']=['A','B']
>>> l
SortedDict(None, 1000, {'A': ['B', 'C'], 'B': ['A', 'C'], 'C': ['D'], 'D': [], 'G': ['A', 'B']})
>>> l['E']=['C','D','G']
>>> l
SortedDict(None, 1000, {'A': ['B', 'C'], 'B': ['A', 'C'], 'C': ['D'], 'D': [], 'E': ['C', 'D', 'G'], 'G': ['A', 'B']})
>>>
I have the following list:
initial_list = [['B', 'D', 'A', 'C', 'E']]
On each element of the list I apply a function and put the results in a dictionary:
for state in initial_list:
next_dict[state] = move([state], alphabet)
This gives the following result:
next_dict = {'D': ['E'], 'B': ['D'], 'A': ['C'], 'C': ['C'], 'E': ['D']}
What I would like to do is separate the keys from initial_list based on their
values in the next_dict dictionary, basically group the elements of the first list to elements with the same value in the next_dict:
new_list = [['A', 'C'], ['B', 'E'], ['D']]
'A' and 'C' will stay in the same group because they have the same value 'C', 'B' and 'D' will also share the same group because their value is 'D' and then 'D' will be in it's own group.
How can I achieve this result?
You need groupby, after having sorted your list by next_dict values :
It generates a break or new group every time the value of the key
function changes (which is why it is usually necessary to have sorted
the data using the same key function).
from itertools import groupby
initial_list = ['B', 'D', 'A', 'C', 'E']
def move(letter):
return {'A': 'C', 'C': 'C', 'D': 'E', 'E': 'D', 'B': 'D'}.get(letter)
sorted_list = sorted(initial_list, key=move)
print [list(v) for k,v in groupby(sorted_list, key=move)]
#=> [['A', 'C'], ['B', 'E'], ['D']]
Simplest way to achieve this will be to use itertools.groupby with key as dict.get as:
>>> from itertools import groupby
>>> next_dict = {'D': ['E'], 'B': ['D'], 'A': ['C'], 'C': ['C'], 'E': ['D']}
>>> initial_list = ['B', 'D', 'A', 'C', 'E']
>>> [list(i) for _, i in groupby(sorted(initial_list, key=next_dict.get), next_dict.get)]
[['A', 'C'], ['B', 'E'], ['D']]
I'm not exactly sure that's what you want but you can group the values based on their values in the next_dict:
>>> next_dict = {'D': 'E', 'B': 'D', 'A': 'C', 'C': 'C', 'E': 'D'}
>>> # external library but one can also use a defaultdict.
>>> from iteration_utilities import groupedby
>>> groupings = groupedby(['B', 'D', 'A', 'C', 'E'], key=next_dict.__getitem__)
>>> groupings
{'C': ['A', 'C'], 'D': ['B', 'E'], 'E': ['D']}
and then convert that to a list of their values:
>>> list(groupings.values())
[['A', 'C'], ['D'], ['B', 'E']]
Combine everything into a one-liner (not really recommended but a lot of people prefer that):
>>> list(groupedby(['B', 'D', 'A', 'C', 'E'], key=next_dict.__getitem__).values())
[['A', 'C'], ['D'], ['B', 'E']]
Try this:
next_next_dict = {}
for key in next_dict:
if next_dict[key][0] in next_next_dict:
next_next_dict[next_dict[key][0]] += key
else:
next_next_dict[next_dict[key][0]] = [key]
new_list = next_next_dict.values()
Or this:
new_list = []
for value in next_dict.values():
new_value = [key for key in next_dict.keys() if next_dict[key] == value]
if new_value not in new_list:
new_list.append(new_value)
We can sort your list with your dictionary mapping, and then use itertools.groupby to form the groups. The only amendment I made here is making your initial list an actual flat list.
>>> from itertools import groupby
>>> initial_list = ['B', 'D', 'A', 'C', 'E']
>>> next_dict = {'D': ['E'], 'B': ['D'], 'A': ['C'], 'C': ['C'], 'E': ['D']}
>>> s_key = lambda x: next_dict[x]
>>> [list(v) for k, v in groupby(sorted(initial_list, key=s_key), key=s_key)]
[['A', 'C'], ['B', 'E'], ['D']]
I have a dictionary that consists of employee-manager as key-value pairs:
{'a': 'b', 'b': 'd', 'c': 'd', 'd': 'f'}
I want to show the relations between employee-manager at all levels (employee's boss, his boss's boss, his boss's boss's boss etc.) using a dictionary. The desired output is:
{'a': [b,d,f], 'b': [d,f], 'c': [d,f], 'd': [f] }
Here is my attempt which only shows the first level:
for key, value in data.items():
if (value in data.keys()):
data[key] = [value]
data[key].append(data[value])
I can do another conditional statement to add the next level but this would be the wrong way to go about it. I'm not very familiar with dictionaries so what would be a better approach?
>>> D = {'a': 'b', 'b': 'd', 'c': 'd', 'd': 'f'}
>>> res = {}
>>> for k in D:
... res[k] = [j] = [D[k]]
... while j in D:
... j = D[j]
... res[k].append(j)
...
>>> res
{'b': ['d', 'f'], 'c': ['d', 'f'], 'd': ['f'], 'a': ['b', 'd', 'f']}
You may use the concept of recursion as :
def get_linked_list(element, hierarchy, lst):
if element:
lst.append(element)
return get_linked_list(hierarchy.get(element, ""), hierarchy, lst)
else:
return lst
And then access the hierarchy as:
>>> d = {'a': 'b', 'b': 'd', 'c': 'd', 'd': 'f'}
>>> print {elem:get_linked_list(elem, d, [])[1:] for elem in d.keys()}
>>> {'a': ['b', 'd', 'f'], 'c': ['d', 'f'], 'b': ['d', 'f'], 'd': ['f']}
However care must be taken as this may get to an infinite loop if we have an item in the dictionary as "a": "a"
x={'a': 'b', 'b': 'd', 'c': 'd', 'd': 'f'}
d={}
l=x.keys()
for i in l:
d.setdefault(i,[])
d[i].append(x[i])
for j in l[l.index(i)+1:]:
if j==d[i][-1]:
d[i].append(x[j])
print d
Output:{'a': ['b', 'd', 'f'], 'c': ['d', 'f'], 'b': ['d', 'f'], 'd': ['f']}
x = [['a', 'b', 'c'], ['a', 'c', 'd'], ['e', 'f', 'f']]
Let's say we have a list with random str letters.
How can i create a function so it tells me how many times the letter 'a' comes out, which in this case 2. Or any other letter, like 'b' comes out once, 'f' comes out twice. etc.
Thank you!
You could flatten the list and use collections.Counter:
>>> import collections
>>> x = [['a', 'b', 'c'], ['a', 'c', 'd'], ['e', 'f', 'f']]
>>> d = collections.Counter(e for sublist in x for e in sublist)
>>> d
Counter({'a': 2, 'c': 2, 'f': 2, 'b': 1, 'e': 1, 'd': 1})
>>> d['a']
2
import itertools, collections
result = collections.defaultdict(int)
for i in itertools.chain(*x):
result[i] += 1
This will create result as a dictionary with the characters as keys and their counts as values.
Just FYI, you can use sum() to flatten a single nested list.
>>> from collections import Counter
>>>
>>> x = [['a', 'b', 'c'], ['a', 'c', 'd'], ['e', 'f', 'f']]
>>> c = Counter(sum(x, []))
>>> c
Counter({'a': 2, 'c': 2, 'f': 2, 'b': 1, 'e': 1, 'd': 1})
But, as Blender and John Clements have addressed, itertools.chain.from_iterable() may be more clear.
>>> from itertools import chain
>>> c = Counter(chain.from_iterable(x)))
>>> c
Counter({'a': 2, 'c': 2, 'f': 2, 'b': 1, 'e': 1, 'd': 1})