问题
I have been working on getting the count(frequency) and then making the graph representation of it.
I am using Counter
class from collections
using python. I want the output of the Counter
in the order of the first come object.
for example:
offset=['a','b','c','a','b','b','b','c','c','c','c','c']
counts = Counter(offset)
print counts
output is:
Counter({'c': 6, 'b': 4, 'a': 2})
As I want the output to be as follows, in the order of the first come object:
Counter({'a': 2, 'b': 4, 'c': 6})
Is that possible?
Thanks
回答1:
OrderedCounter by (2 lines long) multiple inheritance
Following wonderful speech Super considered super by Raymond Hettinger we can use multiple inheritance:
from collections import OrderedDict, Counter
class OrderedCounter(Counter, OrderedDict):
pass
and use it:
>>> counter = OrderedCounter("abracadabra")
>>> for key, value in counter.items():
... print key, value
a 5
b 2
r 2
c 1
d 1
This is all you need for your task.
Fixing a bit confusiong representation
Let us do a bit more tests and see, what "result" we get:
>>> counter = OrderedCounter("cbaaa")
>>> counter
OrderedCounter({'a': 3, 'c': 1, 'b': 1})
Oh - this looks wrong, expected order is "c", "b", "a". Let us test it printing the keys and values:
>>> for key, value in counter.items():
... print key, value
c 1
b 1
a 3
This looks correct (and it is exactly what really counts for using in your code).
It turns out, that the class we have created is just producing a bit confusing representation.
This can be fixed:
class OrderedCounter(Counter, OrderedDict):
def __repr__(self):
return "%s(%r)" % (self.__class__.__name__, OrderedDict(self))
and when used:
>>> counter = OrderedCounter("cbaaa")
>>> counter
OrderedCounter({'c': 1, 'b': 1, 'a': 1})
Full version from the Raymond Hettinger speech (added pickling)
Full version provided in the speech is adding one more method __reduce__
which allows pickling
the object properly.
from collections import OrderedDict, Counter
class OrderedCounter(Counter, OrderedDict):
"""Counter that remembers the order elements are first seen"""
def __repr__(self):
return "%s(%r)" % (self.__class_.__name__, OrderedDict(self))
def __reduce__(self):
return self.__class__, (OrderedDict(self),)
Anyway, in most cases you will manage with the simplest version of the OrderedCounter
class.
回答2:
EDIT
sorry, I misunderstood. try this. I implemented that using queue concept.
q = []
ret = []
for i in offset:
if q.count(i)==0: q.insert(i, 0)
while len(q):
item = q.pop()
ret.append((item, d.get(item)))
print(ret)
回答3:
Dictionaries in python are just hash tables so they don't have any sort of order. You can never print out a sorted list. But you can convert them to a list of tuples and sort that.
from collections import Counter
import operator
offset=['a','b','c','a','b','b','b','c','c','c','c','c']
counts = Counter(offset)
print(sorted(counts.items(), key=operator.itemgetter(0)))
#[('a', 2), ('b', 4), ('c', 6)]
回答4:
You will have to do it "the old fashion way" using an ordered dictionary that guarantees to retain the order in which elements are first inserted:
from collections import OrderedDict
offset = ['a','b','c','a','b','b','b','c','c','c','c','c']
counts = OrderedDict()
for elt in offset:
try:
counts[elt] += 1
except KeyError:
counts[elt] = 1
print(counts)
Result:
OrderedDict([('a', 2), ('b', 4), ('c', 6)])
回答5:
You can use such a methodology:
from collections import Counter
offset=['a','b','c','a','b','b','b','c','c','c','c','c']
counts = Counter(offset)
for letter in offset:
if letter in counts:
print (letter + ": " + str(counts[letter]))
counts.pop(letter)
That works as below:
>>> ================================ RESTART ================================
>>>
a: 2
b: 4
c: 6
You can store the pairs in a list instead of printing.
来源:https://stackoverflow.com/questions/36807701/how-to-get-the-python-counter-output-ordered-by-order-of-inputs