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find all the elements with equal counts with Counter().most_common

I am using Counter().most_common from collections. My input(e.g counter_list) has many equal elements but if i use Counter(mylist).most_common(1) instead of getting all the results i get only the first
mylist=['gene0.txt','gene1.txt','gene1.txt','gene2.txt','gene2.txt','gene3.txt','gene3.txt']
C = Counter(mylist).most_common(1)
I am getting this
[('gene1.txt',2)]
Instead of that
[('gene1.txt',2),('gene2.txt',2),('gene3.txt',2)]

One solution is using itertools.takewhile:
mylist=['gene0.txt','gene1.txt','gene1.txt','gene2.txt','gene2.txt','gene3.txt','gene3.txt']
from collections import Counter
from itertools import takewhile
c = Counter(mylist)
print(list(takewhile(lambda val: val[1] == c.most_common(1)[0][1], c.most_common())))
Prints:
[('gene1.txt', 2), ('gene2.txt', 2), ('gene3.txt', 2)]
Or without itertools.takewhile, using list comprehension:
print([item for item in c.most_common() if item[1] == c.most_common(1)[0][1]])
But this will iterate over all items in most_common().
EDIT (Some debug info for clarity):
for val in c.most_common():
print('val[0]={} val[1]={} c.most_common(1)[0][1]={}'.format(val[0], val[1], c.most_common(1)[0][1]))
Prints:
val[0]=gene1.txt val[1]=2 c.most_common(1)[0][1]=2
val[0]=gene2.txt val[1]=2 c.most_common(1)[0][1]=2
val[0]=gene3.txt val[1]=2 c.most_common(1)[0][1]=2
val[0]=gene0.txt val[1]=1 c.most_common(1)[0][1]=2

bubble sorting in dictionary in python

I am trying to do bubble sort in descending order
input={"m":1,"i":4,"s":4,"P":2}
output={"i":4,"s":4,"p":2,"m":1}
but a type error occurs. How can I do this?
Below is the code:
dict={"m":1,"i":4,"s":4,"P":2}
for key,value in dict.items():
if(dict[key]<dict[key+1]):
temp=dict[key]
dict[key]=dict[key+1]
dict[key+1]=temp
print(dict)

A dictionary has no order: if you add/remove/update elements in a dictionary, the order in which the keys are iterated can change.
So you can not "sort" a dictionary. What you can do however, is for example sort a list of 2-tuples, like:
my_list = list(my_dict.items())
We then retrieve:
>>> my_list
[('m', 1), ('i', 4), ('s', 4), ('P', 2)]
and thnen we can sort the list (for example with bubblesort) like:
for mx in range(len(my_list)-1, -1, -1):
swapped = False
for i in range(mx):
if my_list[i][1] < my_list[i+1][1]:
my_list[i], my_list[i+1] = my_list[i+1], my_list[i]
swapped = True
if not swapped:
break
Then afterwards, we have:
>>> my_list
[('i', 4), ('s', 4), ('P', 2), ('m', 1)]

do you mean something like this?
dict={"m":1,"i":4,"s":4,"P":2}
sorted_values = sorted(dict.values(),reverse = True)
print sorted_values
But this isn't "manual" bubble sort. So if you have to to bubble sort this by your own just use the list of values via
values = dict.values()
And then sort them ;)
As timgeb already mentioned in a comment, I'm sorry didnt see the comment before I wrote this post.

Python group by adjacent items in a list with same attributes

Please see the simplified example:
A=[(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
I want group adjacent tuples with attribute 'a', every two pair wise and leave tuples with 'b' alone.
So the desired tuple would looks like:
A=[[(721,'a'),(765,'a')],
[(421,'a'),(422,'a')],
[(106,'b')],
[(784,'a'),(201,'a')],
[(206,'b')],[(207,'b')]]
What I can do is to build two separated lists contains tuples with a and b.
Then pair tuples in a, and add back. But it seems not very efficient. Any faster and simple solutions?

Assuming a items are always in pairs, a simple approach would be as follows.
Look at the first item - if it's an a, use it and the next item as a pair. Otherwise, just use the single item. Then 'jump' forward by 1 or 2, as appropriate:
A=[(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
result = []
count = 0
while count <= len(A)-1:
if A[count][1] == 'a':
result.append([A[count], A[count+1]])
count += 2
else:
result.append([A[count]])
count += 1
print(result)

You can use itertools.groupby:
import itertools
A=[(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
def split(s):
return [s[i:i+2] for i in range(0, len(s), 2)]
new_data = [i if isinstance(i, list) else [i] for i in list(itertools.chain(*[split(list(b)) if a == 'a' else list(b) for a, b in itertools.groupby(A, key=lambda x:x[-1])]))
Output:
[[(721, 'a'), (765, 'a')], [(421, 'a'), (422, 'a')], [(106, 'b')], [(784, 'a'), (201, 'a')], [(206, 'b')], [(207, 'b')]]

No need to use two lists. Edit: If the 'a' are not assumed to come always as pairs/adjacent
A = [(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
new_list = []
i = 0
while i < len(A):
if i == len(A)-1:
new_list.append([A[i]])
i+=1
elif (A[i][1]==A[i+1][1]=='a') :
new_list.append([A[i], A[i+1]])
i += 2
else:
new_list.append([A[i]])
i += 1
print(new_list)

return the top n most frequently occurring chars and their respective counts in python

how to return the top n most frequently occurring chars and their respective counts # e.g 'aaaaaabbbbcccc', 2 should return [('a', 6), ('b', 4)] in python
I tried this
def top_chars(input, n):
list1=list(input)
list3=[]
list2=[]
list4=[]
set1=set(list1)
list2=list(set1)
def count(item):
count=0
for x in input:
if x in input:
count+=item.count(x)
list3.append(count)
return count
list2.sort(key=count)
list3.sort()
list4=list(zip(list2,list3))
list4.reverse()
list4.sort(key=lambda list4: ((list4[1]),(list4[0])), reverse=True)
return list4[0:n]
pass
but it doesn't work for the input ("aabc",2)
The output it should give is
[('a', 2), ('b', 1)]
but the output I get is
[('a', 2), ('c', 1)]

Use collections.Counter(); it has a most_common() method that does just that:
>>> from collections import Counter
>>> counts = Counter('aaaaaabbbbcccc')
>>> counts.most_common(2)
[('a', 6), ('c', 4)]
Note that for both the above input and in aabc both b and c have the same count, and both can be valid top contenders. Because both you and Counter sort by count then key in reverse, c is sorted before b.
If instead of sorting in reverse, you used the negative count as the sort key, you'd sort b before c again:
list4.sort(key=lambda v: (-v[1], v[0))
Not that Counter.most_common() actually uses sorting when your are asking for fewer items than there are keys in the counter; it uses a heapq-based algorithm instead to only get the top N items.

A little harder, but also works:
text = "abbbaaaa"
dict = {}
for lines in text:
for char in lines:
dict[char] = dict.get(char, 0) + 1
print dict

Text="abbbaaaa"
dict={ }
For lines in text:
For chae in lines:
dict[char]=dict.get(char,0)+1
Print dict

How do I use itertools.groupby()?

I haven't been able to find an understandable explanation of how to actually use Python's itertools.groupby() function. What I'm trying to do is this:
Take a list - in this case, the children of an objectified lxml element
Divide it into groups based on some criteria
Then later iterate over each of these groups separately.
I've reviewed the documentation, but I've had trouble trying to apply them beyond a simple list of numbers.
So, how do I use of itertools.groupby()? Is there another technique I should be using? Pointers to good "prerequisite" reading would also be appreciated.

IMPORTANT NOTE: You have to sort your data first.
The part I didn't get is that in the example construction
groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
groups.append(list(g)) # Store group iterator as a list
uniquekeys.append(k)
k is the current grouping key, and g is an iterator that you can use to iterate over the group defined by that grouping key. In other words, the groupby iterator itself returns iterators.
Here's an example of that, using clearer variable names:
from itertools import groupby
things = [("animal", "bear"), ("animal", "duck"), ("plant", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
for key, group in groupby(things, lambda x: x[0]):
for thing in group:
print("A %s is a %s." % (thing[1], key))
print("")
This will give you the output:
A bear is a animal.
A duck is a animal.
A cactus is a plant.
A speed boat is a vehicle.
A school bus is a vehicle.
In this example, things is a list of tuples where the first item in each tuple is the group the second item belongs to.
The groupby() function takes two arguments: (1) the data to group and (2) the function to group it with.
Here, lambda x: x[0] tells groupby() to use the first item in each tuple as the grouping key.
In the above for statement, groupby returns three (key, group iterator) pairs - once for each unique key. You can use the returned iterator to iterate over each individual item in that group.
Here's a slightly different example with the same data, using a list comprehension:
for key, group in groupby(things, lambda x: x[0]):
listOfThings = " and ".join([thing[1] for thing in group])
print(key + "s: " + listOfThings + ".")
This will give you the output:
animals: bear and duck.
plants: cactus.
vehicles: speed boat and school bus.

itertools.groupby is a tool for grouping items.
From the docs, we glean further what it might do:
# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
groupby objects yield key-group pairs where the group is a generator.
Features
A. Group consecutive items together
B. Group all occurrences of an item, given a sorted iterable
C. Specify how to group items with a key function *
Comparisons
# Define a printer for comparing outputs
>>> def print_groupby(iterable, keyfunc=None):
... for k, g in it.groupby(iterable, keyfunc):
... print("key: '{}'--> group: {}".format(k, list(g)))
# Feature A: group consecutive occurrences
>>> print_groupby("BCAACACAADBBB")
key: 'B'--> group: ['B']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'D'--> group: ['D']
key: 'B'--> group: ['B', 'B', 'B']
# Feature B: group all occurrences
>>> print_groupby(sorted("BCAACACAADBBB"))
key: 'A'--> group: ['A', 'A', 'A', 'A', 'A']
key: 'B'--> group: ['B', 'B', 'B', 'B']
key: 'C'--> group: ['C', 'C', 'C']
key: 'D'--> group: ['D']
# Feature C: group by a key function
>>> # islower = lambda s: s.islower() # equivalent
>>> def islower(s):
... """Return True if a string is lowercase, else False."""
... return s.islower()
>>> print_groupby(sorted("bCAaCacAADBbB"), keyfunc=islower)
key: 'False'--> group: ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'D']
key: 'True'--> group: ['a', 'a', 'b', 'b', 'c']
Uses
Anagrams (see notebook)
Binning
Group odd and even numbers
Group a list by values
Remove duplicate elements
Find indices of repeated elements in an array
Split an array into n-sized chunks
Find corresponding elements between two lists
Compression algorithm (see notebook)/Run Length Encoding
Grouping letters by length, key function (see notebook)
Consecutive values over a threshold (see notebook)
Find ranges of numbers in a list or continuous items (see docs)
Find all related longest sequences
Take consecutive sequences that meet a condition (see related post)
Note: Several of the latter examples derive from Víctor Terrón's PyCon (talk) (Spanish), "Kung Fu at Dawn with Itertools". See also the groupby source code written in C.
* A function where all items are passed through and compared, influencing the result. Other objects with key functions include sorted(), max() and min().
Response
# OP: Yes, you can use `groupby`, e.g.
[do_something(list(g)) for _, g in groupby(lxml_elements, criteria_func)]

The example on the Python docs is quite straightforward:
groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
groups.append(list(g)) # Store group iterator as a list
uniquekeys.append(k)
So in your case, data is a list of nodes, keyfunc is where the logic of your criteria function goes and then groupby() groups the data.
You must be careful to sort the data by the criteria before you call groupby or it won't work. groupby method actually just iterates through a list and whenever the key changes it creates a new group.

A neato trick with groupby is to run length encoding in one line:
[(c,len(list(cgen))) for c,cgen in groupby(some_string)]
will give you a list of 2-tuples where the first element is the char and the 2nd is the number of repetitions.
Edit: Note that this is what separates itertools.groupby from the SQL GROUP BY semantics: itertools doesn't (and in general can't) sort the iterator in advance, so groups with the same "key" aren't merged.

Another example:
for key, igroup in itertools.groupby(xrange(12), lambda x: x // 5):
print key, list(igroup)
results in
0 [0, 1, 2, 3, 4]
1 [5, 6, 7, 8, 9]
2 [10, 11]
Note that igroup is an iterator (a sub-iterator as the documentation calls it).
This is useful for chunking a generator:
def chunker(items, chunk_size):
'''Group items in chunks of chunk_size'''
for _key, group in itertools.groupby(enumerate(items), lambda x: x[0] // chunk_size):
yield (g[1] for g in group)
with open('file.txt') as fobj:
for chunk in chunker(fobj):
process(chunk)
Another example of groupby - when the keys are not sorted. In the following example, items in xx are grouped by values in yy. In this case, one set of zeros is output first, followed by a set of ones, followed again by a set of zeros.
xx = range(10)
yy = [0, 0, 0, 1, 1, 1, 0, 0, 0, 0]
for group in itertools.groupby(iter(xx), lambda x: yy[x]):
print group[0], list(group[1])
Produces:
0 [0, 1, 2]
1 [3, 4, 5]
0 [6, 7, 8, 9]

WARNING:
The syntax list(groupby(...)) won't work the way that you intend. It seems to destroy the internal iterator objects, so using
for x in list(groupby(range(10))):
print(list(x[1]))
will produce:
[]
[]
[]
[]
[]
[]
[]
[]
[]
[9]
Instead, of list(groupby(...)), try [(k, list(g)) for k,g in groupby(...)], or if you use that syntax often,
def groupbylist(*args, **kwargs):
return [(k, list(g)) for k, g in groupby(*args, **kwargs)]
and get access to the groupby functionality while avoiding those pesky (for small data) iterators all together.

I would like to give another example where groupby without sort is not working. Adapted from example by James Sulak
from itertools import groupby
things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
for key, group in groupby(things, lambda x: x[0]):
for thing in group:
print "A %s is a %s." % (thing[1], key)
print " "
output is
A bear is a vehicle.
A duck is a animal.
A cactus is a animal.
A speed boat is a vehicle.
A school bus is a vehicle.
there are two groups with vehicule, whereas one could expect only one group

#CaptSolo, I tried your example, but it didn't work.
from itertools import groupby
[(c,len(list(cs))) for c,cs in groupby('Pedro Manoel')]
Output:
[('P', 1), ('e', 1), ('d', 1), ('r', 1), ('o', 1), (' ', 1), ('M', 1), ('a', 1), ('n', 1), ('o', 1), ('e', 1), ('l', 1)]
As you can see, there are two o's and two e's, but they got into separate groups. That's when I realized you need to sort the list passed to the groupby function. So, the correct usage would be:
name = list('Pedro Manoel')
name.sort()
[(c,len(list(cs))) for c,cs in groupby(name)]
Output:
[(' ', 1), ('M', 1), ('P', 1), ('a', 1), ('d', 1), ('e', 2), ('l', 1), ('n', 1), ('o', 2), ('r', 1)]
Just remembering, if the list is not sorted, the groupby function will not work!

Sorting and groupby
from itertools import groupby
val = [{'name': 'satyajit', 'address': 'btm', 'pin': 560076},
{'name': 'Mukul', 'address': 'Silk board', 'pin': 560078},
{'name': 'Preetam', 'address': 'btm', 'pin': 560076}]
for pin, list_data in groupby(sorted(val, key=lambda k: k['pin']),lambda x: x['pin']):
... print pin
... for rec in list_data:
... print rec
...
o/p:
560076
{'name': 'satyajit', 'pin': 560076, 'address': 'btm'}
{'name': 'Preetam', 'pin': 560076, 'address': 'btm'}
560078
{'name': 'Mukul', 'pin': 560078, 'address': 'Silk board'}

Sadly I don’t think it’s advisable to use itertools.groupby(). It’s just too hard to use safely, and it’s only a handful of lines to write something that works as expected.
def my_group_by(iterable, keyfunc):
"""Because itertools.groupby is tricky to use
The stdlib method requires sorting in advance, and returns iterators not
lists, and those iterators get consumed as you try to use them, throwing
everything off if you try to look at something more than once.
"""
ret = defaultdict(list)
for k in iterable:
ret[keyfunc(k)].append(k)
return dict(ret)
Use it like this:
def first_letter(x):
return x[0]
my_group_by('four score and seven years ago'.split(), first_letter)
to get
{'f': ['four'], 's': ['score', 'seven'], 'a': ['and', 'ago'], 'y': ['years']}

How do I use Python's itertools.groupby()?
You can use groupby to group things to iterate over. You give groupby an iterable, and a optional key function/callable by which to check the items as they come out of the iterable, and it returns an iterator that gives a two-tuple of the result of the key callable and the actual items in another iterable. From the help:
groupby(iterable[, keyfunc]) -> create an iterator which returns
(key, sub-iterator) grouped by each value of key(value).
Here's an example of groupby using a coroutine to group by a count, it uses a key callable (in this case, coroutine.send) to just spit out the count for however many iterations and a grouped sub-iterator of elements:
import itertools
def grouper(iterable, n):
def coroutine(n):
yield # queue up coroutine
for i in itertools.count():
for j in range(n):
yield i
groups = coroutine(n)
next(groups) # queue up coroutine
for c, objs in itertools.groupby(iterable, groups.send):
yield c, list(objs)
# or instead of materializing a list of objs, just:
# return itertools.groupby(iterable, groups.send)
list(grouper(range(10), 3))
prints
[(0, [0, 1, 2]), (1, [3, 4, 5]), (2, [6, 7, 8]), (3, [9])]

This basic implementation helped me understand this function. Hope it helps others as well:
arr = [(1, "A"), (1, "B"), (1, "C"), (2, "D"), (2, "E"), (3, "F")]
for k,g in groupby(arr, lambda x: x[0]):
print("--", k, "--")
for tup in g:
print(tup[1]) # tup[0] == k
-- 1 --
A
B
C
-- 2 --
D
E
-- 3 --
F

One useful example that I came across may be helpful:
from itertools import groupby
#user input
myinput = input()
#creating empty list to store output
myoutput = []
for k,g in groupby(myinput):
myoutput.append((len(list(g)),int(k)))
print(*myoutput)
Sample input: 14445221
Sample output: (1,1) (3,4) (1,5) (2,2) (1,1)

from random import randint
from itertools import groupby
l = [randint(1, 3) for _ in range(20)]
d = {}
for k, g in groupby(l, lambda x: x):
if not d.get(k, None):
d[k] = list(g)
else:
d[k] = d[k] + list(g)
the code above shows how groupby can be used to group a list based on the lambda function/key supplied. The only problem is that the output is not merged, this can be easily resolved using a dictionary.
Example:
l = [2, 1, 2, 3, 1, 3, 2, 1, 3, 3, 1, 3, 2, 3, 1, 2, 1, 3, 2, 3]
after applying groupby the result will be:
for k, g in groupby(l, lambda x:x):
print(k, list(g))
2 [2]
1 [1]
2 [2]
3 [3]
1 [1]
3 [3]
2 [2]
1 [1]
3 [3, 3]
1 [1]
3 [3]
2 [2]
3 [3]
1 [1]
2 [2]
1 [1]
3 [3]
2 [2]
3 [3]
Once a dictionary is used as shown above following result is derived which can be easily iterated over:
{2: [2, 2, 2, 2, 2, 2], 1: [1, 1, 1, 1, 1, 1], 3: [3, 3, 3, 3, 3, 3, 3, 3]}

The key thing to recognize with itertools.groupby is that items are only grouped together as long as they're sequential in the iterable. This is why sorting works, because basically you're rearranging the collection so that all of the items which satisfy callback(item) now appear in the sorted collection sequentially.
That being said, you don't need to sort the list, you just need a collection of key-value pairs, where the value can grow in accordance to each group iterable yielded by groupby. i.e. a dict of lists.
>>> things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
>>> coll = {}
>>> for k, g in itertools.groupby(things, lambda x: x[0]):
... coll.setdefault(k, []).extend(i for _, i in g)
...
{'vehicle': ['bear', 'speed boat', 'school bus'], 'animal': ['duck', 'cactus']}