Related
Stackoverflow, I am once again asking for your help.
I'm aware there are other threads about this but I'll explain what makes my assignment different.
Basically my function would get a list of 0s and 1s, and return all the possible orders for the string. For example for "0111" we will get "0111", "1011", "1101", "1110".
Here's my code:
def permutations(string):
if len(string) == 1:
return [string]
lst = []
for j in range(len(string)):
remaining_elements = ''.join([string[i] for i in range(len(string)) if i != j])
mini_perm = permutations(remaining_elements)
for perm in mini_perm:
new_str = string[j] + perm
if new_str not in lst:
lst.append(new_str)
return lst
The problem is when I run a string like "000000000011" it takes a very long time to process. There is supposed to be a more efficient way to do it because it's just two numbers. So I shouldn't be using the indexes?
Please help me if you can figure out a more efficient say to do this.
(I am allowed to use loops just have to use recursion as well!)
Here is an example for creating permutations with recursion that is more efficient:
def permute(string):
string = list(string)
n = len(string)
# Base conditions
# If length is 0 or 1, there is only 1 permutation
if n in [0, 1]:
return [string]
# If length is 2, then there are only two permutations
# Example: [1,2] and [2,1]
if n == 2:
return [string, string[::-1]]
res = []
# For every number in array, choose 1 number and permute the remaining
# by calling permute recursively
for i in range(n):
permutations = permute(string[:i] + string[i+1:])
for p in permutations:
res.append([''.join(str(n) for n in [string[i]] + p)])
return res
This should also work for permute('000000000011') - hope it helps!
You can also use collections.Counter with a recursive generator function:
from collections import Counter
def permute(d):
counts = Counter(d)
def get_permuations(c, s = []):
if len(s) == sum(counts.values()):
yield ''.join(s)
else:
for a, b in c.items():
for i in range(1, b+1):
yield from get_permuations({**c, a:b - i}, s+([a]*i))
return list(set(get_permuations(counts)))
print(permute("0111"))
print(permute("000000000011"))
Output:
['0111', '1110', '1101', '1011']
['010000100000', '100000000001', '010000001000', '000000100001', '011000000000', '100000000010', '001001000000', '000000011000', '100000001000', '100000100000', '100001000000', '001000100000', '100010000000', '000000001100', '000100000100', '010010000000', '000000000011', '000000100010', '101000000000', '110000000000', '100000010000', '000100001000', '000001001000', '000000000101', '000000100100', '010000000001', '001000000100', '001000000010', '000110000000', '000011000000', '000001100000', '000000110000', '001000000001', '000010001000', '000100100000', '000001000001', '000010000001', '001100000000', '000100000001', '001000001000', '010000000100', '010000010000', '000000010001', '001000010000', '010001000000', '100000000100', '100100000000', '000000001001', '010100000000', '000010100000', '010000000010', '000000001010', '000010000100', '001010000000', '000000010010', '000001000010', '000100000010', '000101000000', '000000010100', '000100010000', '000000000110', '000001000100', '000010010000', '000000101000', '000001010000', '000010000010']
posting an answer someone gave me. Thanks for your responses!:
def permutations(zeroes, ones, lst, perm):
if zeroes == 0 and ones == 0:
lst.append(perm)
return
elif zeroes < 0 or ones < 0:
return
permutations(zeroes - 1, ones, lst, perm + '0')
permutations(zeroes, ones - 1, lst, perm + '1')
I want to accomplish this:
Construct a function that takes in a list as a parameter and returns the biggest even number in that list.
Do this by using the "fold" function in Python
I thought it might be something along the lines of:
def fold(f, v, l):
for x in l:
v = f(v, x)
return v
def biggest_even_number(xs):
l = [i for i in xs if i % 2 == 0]
return fold(l)
I know this is wrong but I just don't know how to set this up. How would I accomplish the above task using the "fold" function?
fold function looks good. You just need to call it with correct arguments:
def biggest_even_number(xs):
l = [i for i in xs if i % 2 == 0]
return fold(max, float("-inf"), l)
If it is not a homework, you can use builtin reduce() which basically does the same thing:
def biggest_even_number(xs):
l = [i for i in xs if i % 2 == 0]
return reduce(max, l, float("-inf"))
Thanks to #Steven Rumbalski, for anyone trying to find the maximum value of a sequence, you don't even need reduce:
def biggest_even_number(xs):
return max(i for i in xs if i % 2 == 0)
Do something like the following:
def fold(l):
biggest = float("-inf")
for i in l:
biggest = max(i, biggest)
return biggest
def biggest_even_number(xs):
l = [i for i in xs if i % 2 == 0]
return fold(l)
I'm trying to solve this problem on the easy section of coderbyte and the prompt is:
Have the function ArrayAdditionI(arr) take the array of numbers stored in arr and return the string true if any combination of numbers in the array can be added up to equal the largest number in the array, otherwise return the string false. For example: if arr contains [4, 6, 23, 10, 1, 3] the output should return true because 4 + 6 + 10 + 3 = 23. The array will not be empty, will not contain all the same elements, and may contain negative numbers.
Here's my solution.
def ArrayAddition(arr):
arr = sorted(arr, reverse=True)
large = arr.pop(0)
storage = 0
placeholder = 0
for r in range(len(arr)):
for n in arr:
if n + storage == large: return True
elif n + storage < large: storage += n
else: continue
storage = 0
if placeholder == 0: placeholder = arr.pop(0)
else: arr.append(placeholder); placeholder = arr.pop(0)
return False
print ArrayAddition([2,95,96,97,98,99,100])
I'm not even sure if this is correct, but it seems to cover all the numbers I plug in. I'm wondering if there is a better way to solve this through algorithm which I know nothing of. I'm thinking a for within a for within a for, etc loop would do the trick, but I don't know how to do that.
What I have in mind is accomplishing this with A+B, A+C, A+D ... A+B+C ... A+B+C+D+E
e.g)
for i in range(len(arr):
print "III: III{}III".format(i)
storage = []
for j in range(len(arr):
print "JJ: II({}),JJ({})".format(i,j)
for k in range(len(arr):
print "K: I{}, J{}, K{}".format(i,j,k)
I've searched all over and found the suggestion of itertool, but I'm wondering if there is a way to write this code up more raw.
Thanks.
A recursive solution:
def GetSum(n, arr):
if len(arr) == 0 and n != 0:
return False
return (n == 0 or
GetSum(n, arr[1:]) or
GetSum(n-arr[0], arr[1:]))
def ArrayAddition(arr):
arrs = sorted(arr)
return GetSum(arrs[-1], arrs[:-1])
print ArrayAddition([2,95,96,97,98,99,100])
The GetSum function returns False when the required sum is non-zero and there are no items in the array. Then it checks for 3 cases:
If the required sum, n, is zero then the goal is achieved.
If we can get the sum with the remaining items after the first item is removed, then the goal is achieved.
If we can get the required sum minus the first element of the list on the rest of the list the goal is achieved.
Your solution doesn't work.
>>> ArrayAddition([10, 11, 20, 21, 30, 31, 60])
False
The simple solution is to use itertools to iterate over all subsets of the input (that don't contain the largest number):
def subsetsum(l):
l = list(l)
target = max(l)
l.remove(l)
for subset_size in xrange(1+len(l)):
for subset in itertools.combinations(l, subset_size):
if sum(subset) == target:
return True
return False
If you want to avoid itertools, you'll need to generate subsets directly. That can be accomplished by counting in binary and using the set bits to determine which elements to pick:
def subsetsum(l):
l = list(l)
target = max(l)
l.remove(l)
for subset_index in xrange(2**len(l)):
subtotal = 0
for i, num in enumerate(l):
# If bit i is set in subset_index
if subset_index & (1 << i):
subtotal += num
if subtotal == target:
return True
return False
Update: I forgot that you want to check all possible combinations. Use this instead:
def ArrayAddition(l):
for length in range(2, len(l)):
for lst in itertools.combinations(l, length):
if sum(lst) in l:
print(lst, sum(lst))
return True
return False
One-liner solution:
>>> any(any(sum(lst) in l for lst in itertools.combinations(l, length)) for length in range(2, len(l)))
Hope this helps!
Generate all the sums of the powerset and test them against the max
def ArrayAddition(L):
return any(sum(k for j,k in enumerate(L) if 1<<j&i)==max(L) for i in range(1<<len(L)))
You could improve this by doing some preprocessing - find the max first and remove it from L
One more way to do it...
Code:
import itertools
def func(l):
m = max(l)
rem = [itertools.combinations([x for x in l if not x == m],i) for i in range(2,len(l)-1)]
print [item for i in rem for item in i if sum(item)==m ]
if __name__=='__main__':
func([1,2,3,4,5])
Output:
[(1, 4), (2, 3)]
Hope this helps.. :)
If I understood the question correctly, simply this should return what you want:
2*max(a)<=sum(a)
This is a part of my homework assignment and im close to the final answer but not quite yet. I need to write a function that counts odd numbers in a list.
Create a recursive function count_odd(l) which takes as its only argument a list of integers. The function will return a count of the number of list elements that are odd, i.e., not evenly divisible by 2.\
>>> print count_odd([])
0
>>> print count_odd([1, 3, 5])
3
>>> print count_odd([2, 4, 6])
0
>>> print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
8
Here is what i have so far:
#- recursive function count_odd -#
def count_odd(l):
"""returns a count of the odd integers in l.
PRE: l is a list of integers.
POST: l is unchanged."""
count_odd=0
while count_odd<len(l):
if l[count_odd]%2==0:
count_odd=count_odd
else:
l[count_odd]%2!=0
count_odd=count_odd+1
return count_odd
#- test harness
print count_odd([])
print count_odd([1, 3, 5])
print count_odd([2, 4, 6])
print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
Can u help explain what im missing. The first two test harness works fine but i cant get the final two. Thanks!
Since this is homework, consider this pseudo-code that just counts a list:
function count (LIST)
if LIST has more items
// recursive case.
// Add one for the current item we are counting,
// and call count() again to process the *remaining* items.
remaining = everything in LIST except the first item
return 1 + count(remaining)
else
// base case -- what "ends" the recursion
// If an item is removed each time, the list will eventually be empty.
return 0
This is very similar to what the homework is asking for, but it needs to be translate to Python and you must work out the correct recursive case logic.
Happy coding.
def count_odd(L):
return (L[0]%2) + count_odd(L[1:]) if L else 0
Are slices ok? Doesn't feel recursive to me, but I guess the whole thing is kind of against usual idioms (i.e. - recursion of this sort in Python):
def countOdd(l):
if l == list(): return 0 # base case, empty list means we're done
return l[0] % 2 + countOdd(l[1:]) # add 1 (or don't) depending on odd/even of element 0. recurse on the rest
x%2 is 1 for odds, 0 for evens. If you are uncomfortable with it or just don't understand it, use the following in place of the last line above:
thisElement = l[0]
restOfList = l[1:]
if thisElement % 2 == 0: currentElementOdd = 0
else: currentElementOdd = 1
return currentElementOdd + countOdd(restOfList)
PS - this is pretty recursive, see what your teacher says if you turn this in =P
>>> def countOdd(l):
... return fold(lambda x,y: x+(y&1),l,0)
...
>>> def fold(f,l,a):
... if l == list(): return a
... return fold(f,l[1:],f(a,l[0]))
All of the prior answers are subdividing the problem into subproblems of size 1 and size n-1. Several people noted that the recursive stack might easily blow out. This solution should keep the recursive stack size at O(log n):
def count_odd(series):
l = len(series) >> 1
if l < 1:
return series[0] & 1 if series else 0
else:
return count_odd(series[:l]) + count_odd(series[l:])
The goal of recursion is to divide the problem into smaller pieces, and apply the solution to the smaller pieces. In this case, we can check if the first number of the list (l[0]) is odd, then call the function again (this is the "recursion") with the rest of the list (l[1:]), adding our current result to the result of the recursion.
def count_odd(series):
if not series:
return 0
else:
left, right = series[0], series[1:]
return count_odd(right) + (1 if (left & 1) else 0)
Tail recursion
def count_odd(integers):
def iter_(lst, count):
return iter_(rest(lst), count + is_odd(first(lst))) if lst else count
return iter_(integers, 0)
def is_odd(integer):
"""Whether the `integer` is odd."""
return integer % 2 != 0 # or `return integer & 1`
def first(lst):
"""Get the first element from the `lst` list.
Return `None` if there are no elements.
"""
return lst[0] if lst else None
def rest(lst):
"""Return `lst` list without the first element."""
return lst[1:]
There is no tail-call optimization in Python, so the above version is purely educational.
The call could be visualize as:
count_odd([1,2,3]) # returns
iter_([1,2,3], 0) # could be replaced by; depth=1
iter_([2,3], 0 + is_odd(1)) if [1,2,3] else 0 # `bool([1,2,3])` is True in Python
iter_([2,3], 0 + True) # `True == 1` in Python
iter_([2,3], 1) # depth=2
iter_([3], 1 + is_odd(2)) if [2,3] else 1
iter_([3], 1 + False) # `False == 0` in Python
iter_([3], 1) # depth=3
iter_([], 1 + is_odd(3)) if [3] else 1
iter_([], 2) # depth=4
iter_(rest([]), 2 + is_odd(first([])) if [] else 2 # bool([]) is False in Python
2 # the answer
Simple trampolining
To avoid 'max recursion depth exceeded' errors for large arrays all tail calls in recursive functions can be wrapped in lambda: expressions; and special trampoline() function can be used to unwrap such expressions. It effectively converts recursion into iterating over a simple loop:
import functools
def trampoline(function):
"""Resolve delayed calls."""
#functools.wraps(function)
def wrapper(*args):
f = function(*args)
while callable(f):
f = f()
return f
return wrapper
def iter_(lst, count):
#NOTE: added `lambda:` before the tail call
return (lambda:iter_(rest(lst), count+is_odd(first(lst)))) if lst else count
#trampoline
def count_odd(integers):
return iter_(integers, 0)
Example:
count_odd([1,2,3])
iter_([1,2,3], 0) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([2,3], 0+is_odd(1)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([3], 1+is_odd(2)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([], 1+is_odd(3))
2 # callable(2) is False
I would write it like this:
def countOddNumbers(numbers):
sum = 0
for num in numbers:
if num%2!=0:
sum += numbers.count(num)
return sum
not sure if i got your question , but as above something similar:
def countOddNumbers(numbers):
count=0
for i in numbers:
if i%2!=0:
count+=1
return count
Generator can give quick result in one line code:
sum((x%2 for x in nums))
What is an efficient way to find the most common element in a Python list?
My list items may not be hashable so can't use a dictionary.
Also in case of draws the item with the lowest index should be returned. Example:
>>> most_common(['duck', 'duck', 'goose'])
'duck'
>>> most_common(['goose', 'duck', 'duck', 'goose'])
'goose'
A simpler one-liner:
def most_common(lst):
return max(set(lst), key=lst.count)
Borrowing from here, this can be used with Python 2.7:
from collections import Counter
def Most_Common(lst):
data = Counter(lst)
return data.most_common(1)[0][0]
Works around 4-6 times faster than Alex's solutions, and is 50 times faster than the one-liner proposed by newacct.
On CPython 3.6+ (any Python 3.7+) the above will select the first seen element in case of ties. If you're running on older Python, to retrieve the element that occurs first in the list in case of ties you need to do two passes to preserve order:
# Only needed pre-3.6!
def most_common(lst):
data = Counter(lst)
return max(lst, key=data.get)
With so many solutions proposed, I'm amazed nobody's proposed what I'd consider an obvious one (for non-hashable but comparable elements) -- [itertools.groupby][1]. itertools offers fast, reusable functionality, and lets you delegate some tricky logic to well-tested standard library components. Consider for example:
import itertools
import operator
def most_common(L):
# get an iterable of (item, iterable) pairs
SL = sorted((x, i) for i, x in enumerate(L))
# print 'SL:', SL
groups = itertools.groupby(SL, key=operator.itemgetter(0))
# auxiliary function to get "quality" for an item
def _auxfun(g):
item, iterable = g
count = 0
min_index = len(L)
for _, where in iterable:
count += 1
min_index = min(min_index, where)
# print 'item %r, count %r, minind %r' % (item, count, min_index)
return count, -min_index
# pick the highest-count/earliest item
return max(groups, key=_auxfun)[0]
This could be written more concisely, of course, but I'm aiming for maximal clarity. The two print statements can be uncommented to better see the machinery in action; for example, with prints uncommented:
print most_common(['goose', 'duck', 'duck', 'goose'])
emits:
SL: [('duck', 1), ('duck', 2), ('goose', 0), ('goose', 3)]
item 'duck', count 2, minind 1
item 'goose', count 2, minind 0
goose
As you see, SL is a list of pairs, each pair an item followed by the item's index in the original list (to implement the key condition that, if the "most common" items with the same highest count are > 1, the result must be the earliest-occurring one).
groupby groups by the item only (via operator.itemgetter). The auxiliary function, called once per grouping during the max computation, receives and internally unpacks a group - a tuple with two items (item, iterable) where the iterable's items are also two-item tuples, (item, original index) [[the items of SL]].
Then the auxiliary function uses a loop to determine both the count of entries in the group's iterable, and the minimum original index; it returns those as combined "quality key", with the min index sign-changed so the max operation will consider "better" those items that occurred earlier in the original list.
This code could be much simpler if it worried a little less about big-O issues in time and space, e.g....:
def most_common(L):
groups = itertools.groupby(sorted(L))
def _auxfun((item, iterable)):
return len(list(iterable)), -L.index(item)
return max(groups, key=_auxfun)[0]
same basic idea, just expressed more simply and compactly... but, alas, an extra O(N) auxiliary space (to embody the groups' iterables to lists) and O(N squared) time (to get the L.index of every item). While premature optimization is the root of all evil in programming, deliberately picking an O(N squared) approach when an O(N log N) one is available just goes too much against the grain of scalability!-)
Finally, for those who prefer "oneliners" to clarity and performance, a bonus 1-liner version with suitably mangled names:-).
from itertools import groupby as g
def most_common_oneliner(L):
return max(g(sorted(L)), key=lambda(x, v):(len(list(v)),-L.index(x)))[0]
What you want is known in statistics as mode, and Python of course has a built-in function to do exactly that for you:
>>> from statistics import mode
>>> mode([1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 6])
3
Note that if there is no "most common element" such as cases where the top two are tied, this will raise StatisticsError on Python
<=3.7, and on 3.8 onwards it will return the first one encountered.
Without the requirement about the lowest index, you can use collections.Counter for this:
from collections import Counter
a = [1936, 2401, 2916, 4761, 9216, 9216, 9604, 9801]
c = Counter(a)
print(c.most_common(1)) # the one most common element... 2 would mean the 2 most common
[(9216, 2)] # a set containing the element, and it's count in 'a'
If they are not hashable, you can sort them and do a single loop over the result counting the items (identical items will be next to each other). But it might be faster to make them hashable and use a dict.
def most_common(lst):
cur_length = 0
max_length = 0
cur_i = 0
max_i = 0
cur_item = None
max_item = None
for i, item in sorted(enumerate(lst), key=lambda x: x[1]):
if cur_item is None or cur_item != item:
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
max_length = cur_length
max_i = cur_i
max_item = cur_item
cur_length = 1
cur_i = i
cur_item = item
else:
cur_length += 1
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
return cur_item
return max_item
This is an O(n) solution.
mydict = {}
cnt, itm = 0, ''
for item in reversed(lst):
mydict[item] = mydict.get(item, 0) + 1
if mydict[item] >= cnt :
cnt, itm = mydict[item], item
print itm
(reversed is used to make sure that it returns the lowest index item)
Sort a copy of the list and find the longest run. You can decorate the list before sorting it with the index of each element, and then choose the run that starts with the lowest index in the case of a tie.
A one-liner:
def most_common (lst):
return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a[1])[0]
I am doing this using scipy stat module and lambda:
import scipy.stats
lst = [1,2,3,4,5,6,7,5]
most_freq_val = lambda x: scipy.stats.mode(x)[0][0]
print(most_freq_val(lst))
Result:
most_freq_val = 5
# use Decorate, Sort, Undecorate to solve the problem
def most_common(iterable):
# Make a list with tuples: (item, index)
# The index will be used later to break ties for most common item.
lst = [(x, i) for i, x in enumerate(iterable)]
lst.sort()
# lst_final will also be a list of tuples: (count, index, item)
# Sorting on this list will find us the most common item, and the index
# will break ties so the one listed first wins. Count is negative so
# largest count will have lowest value and sort first.
lst_final = []
# Get an iterator for our new list...
itr = iter(lst)
# ...and pop the first tuple off. Setup current state vars for loop.
count = 1
tup = next(itr)
x_cur, i_cur = tup
# Loop over sorted list of tuples, counting occurrences of item.
for tup in itr:
# Same item again?
if x_cur == tup[0]:
# Yes, same item; increment count
count += 1
else:
# No, new item, so write previous current item to lst_final...
t = (-count, i_cur, x_cur)
lst_final.append(t)
# ...and reset current state vars for loop.
x_cur, i_cur = tup
count = 1
# Write final item after loop ends
t = (-count, i_cur, x_cur)
lst_final.append(t)
lst_final.sort()
answer = lst_final[0][2]
return answer
print most_common(['x', 'e', 'a', 'e', 'a', 'e', 'e']) # prints 'e'
print most_common(['goose', 'duck', 'duck', 'goose']) # prints 'goose'
Building on Luiz's answer, but satisfying the "in case of draws the item with the lowest index should be returned" condition:
from statistics import mode, StatisticsError
def most_common(l):
try:
return mode(l)
except StatisticsError as e:
# will only return the first element if no unique mode found
if 'no unique mode' in e.args[0]:
return l[0]
# this is for "StatisticsError: no mode for empty data"
# after calling mode([])
raise
Example:
>>> most_common(['a', 'b', 'b'])
'b'
>>> most_common([1, 2])
1
>>> most_common([])
StatisticsError: no mode for empty data
Simple one line solution
moc= max([(lst.count(chr),chr) for chr in set(lst)])
It will return most frequent element with its frequency.
You probably don't need this anymore, but this is what I did for a similar problem. (It looks longer than it is because of the comments.)
itemList = ['hi', 'hi', 'hello', 'bye']
counter = {}
maxItemCount = 0
for item in itemList:
try:
# Referencing this will cause a KeyError exception
# if it doesn't already exist
counter[item]
# ... meaning if we get this far it didn't happen so
# we'll increment
counter[item] += 1
except KeyError:
# If we got a KeyError we need to create the
# dictionary key
counter[item] = 1
# Keep overwriting maxItemCount with the latest number,
# if it's higher than the existing itemCount
if counter[item] > maxItemCount:
maxItemCount = counter[item]
mostPopularItem = item
print mostPopularItem
ans = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)
all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())
4
print(all_ans[max_key])
1
#This will return the list sorted by frequency:
def orderByFrequency(list):
listUniqueValues = np.unique(list)
listQty = []
listOrderedByFrequency = []
for i in range(len(listUniqueValues)):
listQty.append(list.count(listUniqueValues[i]))
for i in range(len(listQty)):
index_bigger = np.argmax(listQty)
for j in range(listQty[index_bigger]):
listOrderedByFrequency.append(listUniqueValues[index_bigger])
listQty[index_bigger] = -1
return listOrderedByFrequency
#And this will return a list with the most frequent values in a list:
def getMostFrequentValues(list):
if (len(list) <= 1):
return list
list_most_frequent = []
list_ordered_by_frequency = orderByFrequency(list)
list_most_frequent.append(list_ordered_by_frequency[0])
frequency = list_ordered_by_frequency.count(list_ordered_by_frequency[0])
index = 0
while(index < len(list_ordered_by_frequency)):
index = index + frequency
if(index < len(list_ordered_by_frequency)):
testValue = list_ordered_by_frequency[index]
testValueFrequency = list_ordered_by_frequency.count(testValue)
if (testValueFrequency == frequency):
list_most_frequent.append(testValue)
else:
break
return list_most_frequent
#tests:
print(getMostFrequentValues([]))
print(getMostFrequentValues([1]))
print(getMostFrequentValues([1,1]))
print(getMostFrequentValues([2,1]))
print(getMostFrequentValues([2,2,1]))
print(getMostFrequentValues([1,2,1,2]))
print(getMostFrequentValues([1,2,1,2,2]))
print(getMostFrequentValues([3,2,3,5,6,3,2,2]))
print(getMostFrequentValues([1,2,2,60,50,3,3,50,3,4,50,4,4,60,60]))
Results:
[]
[1]
[1]
[1, 2]
[2]
[1, 2]
[2]
[2, 3]
[3, 4, 50, 60]
Here:
def most_common(l):
max = 0
maxitem = None
for x in set(l):
count = l.count(x)
if count > max:
max = count
maxitem = x
return maxitem
I have a vague feeling there is a method somewhere in the standard library that will give you the count of each element, but I can't find it.
This is the obvious slow solution (O(n^2)) if neither sorting nor hashing is feasible, but equality comparison (==) is available:
def most_common(items):
if not items:
raise ValueError
fitems = []
best_idx = 0
for item in items:
item_missing = True
i = 0
for fitem in fitems:
if fitem[0] == item:
fitem[1] += 1
d = fitem[1] - fitems[best_idx][1]
if d > 0 or (d == 0 and fitems[best_idx][2] > fitem[2]):
best_idx = i
item_missing = False
break
i += 1
if item_missing:
fitems.append([item, 1, i])
return items[best_idx]
But making your items hashable or sortable (as recommended by other answers) would almost always make finding the most common element faster if the length of your list (n) is large. O(n) on average with hashing, and O(n*log(n)) at worst for sorting.
>>> li = ['goose', 'duck', 'duck']
>>> def foo(li):
st = set(li)
mx = -1
for each in st:
temp = li.count(each):
if mx < temp:
mx = temp
h = each
return h
>>> foo(li)
'duck'
I needed to do this in a recent program. I'll admit it, I couldn't understand Alex's answer, so this is what I ended up with.
def mostPopular(l):
mpEl=None
mpIndex=0
mpCount=0
curEl=None
curCount=0
for i, el in sorted(enumerate(l), key=lambda x: (x[1], x[0]), reverse=True):
curCount=curCount+1 if el==curEl else 1
curEl=el
if curCount>mpCount \
or (curCount==mpCount and i<mpIndex):
mpEl=curEl
mpIndex=i
mpCount=curCount
return mpEl, mpCount, mpIndex
I timed it against Alex's solution and it's about 10-15% faster for short lists, but once you go over 100 elements or more (tested up to 200000) it's about 20% slower.
def most_frequent(List):
counter = 0
num = List[0]
for i in List:
curr_frequency = List.count(i)
if(curr_frequency> counter):
counter = curr_frequency
num = i
return num
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
Hi this is a very simple solution, with linear time complexity
L = ['goose', 'duck', 'duck']
def most_common(L):
current_winner = 0
max_repeated = None
for i in L:
amount_times = L.count(i)
if amount_times > current_winner:
current_winner = amount_times
max_repeated = i
return max_repeated
print(most_common(L))
"duck"
Where number, is the element in the list that repeats most of the time
numbers = [1, 3, 7, 4, 3, 0, 3, 6, 3]
max_repeat_num = max(numbers, key=numbers.count) *# which number most* frequently
max_repeat = numbers.count(max_repeat_num) *#how many times*
print(f" the number {max_repeat_num} is repeated{max_repeat} times")
def mostCommonElement(list):
count = {} // dict holder
max = 0 // keep track of the count by key
result = None // holder when count is greater than max
for i in list:
if i not in count:
count[i] = 1
else:
count[i] += 1
if count[i] > max:
max = count[i]
result = i
return result
mostCommonElement(["a","b","a","c"]) -> "a"
The most common element should be the one which is appearing more than N/2 times in the array where N being the len(array). The below technique will do it in O(n) time complexity, with just consuming O(1) auxiliary space.
from collections import Counter
def majorityElement(arr):
majority_elem = Counter(arr)
size = len(arr)
for key, val in majority_elem.items():
if val > size/2:
return key
return -1
def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)
def popular(L):
C={}
for a in L:
C[a]=L.count(a)
for b in C.keys():
if C[b]==max(C.values()):
return b
L=[2,3,5,3,6,3,6,3,6,3,7,467,4,7,4]
print popular(L)