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Sum of random list numbers after 1st negative number

import random
def mainlist(list, size, min, max):
for i in range(size):
list.append(random.randint(min, max))
print(list)
def counterlist(list):
for i in list:
if i<0:
x=sum(list[(list.index(i)+1):])
print('Reqemlerin cemi:', x)
break
list = []
mainlist(list, 10, -10, 30)
counterlist(list)
I need to calculate sum of numbers after 1st negative number in this random list, did it in second function but want to know is there way not using the sum() function?

Explicitly using an iterator makes it nicer and more efficient:
def counterlist(lst):
it = iter(lst)
for i in it:
if i < 0:
print('Reqemlerin cemi:', sum(it))
No idea why you wouldn't want to use the sum function, that's absolutely the right and best way to do it.

Try this:
import random
lst = [random.randint(-10, 30) for _ in range(10)]
print(sum(lst[next(i for i, n in enumerate(lst) if n < 0) + 1:]))
First you generate the list lst. Then, you iterate over your list and you find the first negative element with next(i for i, n in enumerate(lst) if n < 0). Finally, you compute the sum of the portion of the list you're interested about.
If you really don't want to use sum but keep things concise (and you're using python >= 3.8):
import random
lst = [random.randint(-10, 30) for _ in range(10)]
s = 0
print([s := s + x for x in lst[next(i for i, n in enumerate(lst) if n < 0) + 1:]][-1])

Assuming there's a negative value in the list, and with a test list "a":
a = [1,2,3,-7,2,3,4,-1,23,3]
sum(a[(a.index([i for i in a if i < 0][0]) + 1):])
Evaluates to 34 as expected. Could also add a try/except IndexError with a simple sum to catch if there's no negative value.
Edit: updated the index for the search.

Yes, you can iterate over the elements of the list and keep adding them to some var which would store your result. But what for? sum approach is much more clear and python-ish.
Also, don't use list as a list name, it's a reserved word.
# After you find a first negative number (at i position)
j = i + 1
elements_sum = 0
while j < len(list):
elements_sum += list[j]
j += 1

Not as good as the marked answer, but just to know how to make use of numpy, being sure there is a negative number in the list.
Sample list: lst = [12, 2, -3, 4, 5, 10, 100]
You can get your result using np.cumsum:
import numpy as np
np_lst = np.array(lst)
cum_sum = np.cumsum(np_lst)
res = cum_sum[-1] - cum_sum[np_lst<0][0]
res #=> 119

First of all don't use list as a variable name, it's a reserved keyword. Secondly, make your loop as follows:
for index, x in enumerate(list_):
if x < 0:
sum_ = sum(list_[(index + 1):])
print('Reqemlerin cemi:', sum_)
break
That way, you don't need to find a value.
At last if you don't want to use sum
found_negative = false
sum_ = 0
for x in list_:
if found_negative:
sum_ += x
elif x < 0:
found_negative = true
print('Reqemlerin cemi:', sum_)

a function that returns number of sums of a certain number.py

I need to write a function that returns the number of ways of reaching a certain number by adding numbers of a list. For example:
print(p([3,5,8,9,11,12,20], 20))
should return:5
The code I wrote is:
def pow(lis):
power = [[]]
for lst in lis:
for po in power:
power = power + [list(po)+[lst]]
return power
def p(lst, n):
counter1 = 0
counter2 = 0
power_list = pow(lst)
print(power_list)
for p in power_list:
for j in p:
counter1 += j
if counter1 == n:
counter2 += 1
counter1 == 0
else:
counter1 == 0
return counter2
pow() is a function that returns all of the subsets of the list and p should return the number of ways to reach the number n. I keep getting an output of zero and I don't understand why. I would love to hear your input for this.
Thanks in advance.

There are two typos in your code: counter1 == 0 is a boolean, it does not reset anything.
This version should work:
def p(lst, n):
counter2 = 0
power_list = pow(lst)
for p in power_list:
counter1 = 0 #reset the counter for every new subset
for j in p:
counter1 += j
if counter1 == n:
counter2 += 1
return counter2

As tobias_k and Faibbus mentioned, you have a typo: counter1 == 0 instead of counter1 = 0, in two places. The counter1 == 0 produces a boolean object of True or False, but since you don't assign the result of that expression the result gets thrown away. It doesn't raise a SyntaxError, since an expression that isn't assigned is legal Python.
As John Coleman and B. M. mention it's not efficient to create the full powerset and then test each subset to see if it has the correct sum. This approach is ok if the input sequence is small, but it's very slow for even moderately sized sequences, and if you actually create a list containing the subsets rather than using a generator and testing the subsets as they're yielded you'll soon run out of RAM.
B. M.'s first solution is quite efficient since it doesn't produce subsets that are larger than the target sum. (I'm not sure what B. M. is doing with that dict-based solution...).
But we can enhance that approach by sorting the list of sums. That way we can break out of the inner for loop as soon as we detect a sum that's too high. True, we need to sort the sums list on each iteration of the outer for loop, but fortunately Python's TimSort is very efficient, and it's optimized to handle sorting a list that contains sorted sub-sequences, so it's ideal for this application.
def subset_sums(seq, goal):
sums = [0]
for x in seq:
subgoal = goal - x
temp = []
for y in sums:
if y > subgoal:
break
temp.append(y + x)
sums.extend(temp)
sums.sort()
return sum(1 for y in sums if y == goal)
# test
lst = [3, 5, 8, 9, 11, 12, 20]
total = 20
print(subset_sums(lst, total))
lst = range(1, 41)
total = 70
print(subset_sums(lst, total))
output
5
28188
With lst = range(1, 41) and total = 70, this code is around 3 times faster than the B.M. lists version.

A one pass solution with one counter, which minimize additions.
def one_pass_sum(L,target):
sums = [0]
cnt = 0
for x in L:
for y in sums[:]:
z = x+y
if z <= target :
sums.append(z)
if z == target : cnt += 1
return cnt
This way if n=len(L), you make less than 2^n additions against n/2 * 2^n by calculating all the sums.
EDIT :
A more efficient solution, that just counts ways. The idea is to see that if there is k ways to make z-x, there is k more way to do z when x arise.
def enhanced_sum_with_lists(L,target):
cnt=[1]+[0]*target # 1 way to make 0
for x in L:
for z in range(target,x-1,-1): # [target, ..., x+1, x]
cnt[z] += cnt[z-x]
return cnt[target]
But order is important : z must be considered descendant here, to have the good counts (Thanks to PM 2Ring).
This can be very fast (n*target additions) for big lists.
For example :
>>> enhanced_sum_with_lists(range(1,100),2500)
875274644371694133420180815
is obtained in 61 ms. It will take the age of the universe to compute it by the first method.

from itertools import chain, combinations
def powerset_generator(i):
for subset in chain.from_iterable(combinations(i, r) for r in range(len(i)+1)):
yield set(subset)
def count_sum(s, cnt):
return sum(1 for i in powerset_generator(s) if sum(k for k in i) == cnt)
print(count_sum(set([3,5,8,9,11,12,20]), 20))

How to see if the list contains consecutive numbers

I want to test if a list contains consecutive integers and no repetition of numbers.
For example, if I have
l = [1, 3, 5, 2, 4, 6]
It should return True.
How should I check if the list contains up to n consecutive numbers without modifying the original list?
I thought about copying the list and removing each number that appears in the original list and if the list is empty then it will return True.
Is there a better way to do this?

For the whole list, it should just be as simple as
sorted(l) == list(range(min(l), max(l)+1))
This preserves the original list, but making a copy (and then sorting) may be expensive if your list is particularly long.
Note that in Python 2 you could simply use the below because range returned a list object. In 3.x and higher the function has been changed to return a range object, so an explicit conversion to list is needed before comparing to sorted(l)
sorted(l) == range(min(l), max(l)+1))
To check if n entries are consecutive and non-repeating, it gets a little more complicated:
def check(n, l):
subs = [l[i:i+n] for i in range(len(l)) if len(l[i:i+n]) == n]
return any([(sorted(sub) in range(min(l), max(l)+1)) for sub in subs])

The first code removes duplicates but keeps order:
from itertools import groupby, count
l = [1,2,4,5,2,1,5,6,5,3,5,5]
def remove_duplicates(values):
output = []
seen = set()
for value in values:
if value not in seen:
output.append(value)
seen.add(value)
return output
l = remove_duplicates(l) # output = [1, 2, 4, 5, 6, 3]
The next set is to identify which ones are in order, taken from here:
def as_range(iterable):
l = list(iterable)
if len(l) > 1:
return '{0}-{1}'.format(l[0], l[-1])
else:
return '{0}'.format(l[0])
l = ','.join(as_range(g) for _, g in groupby(l, key=lambda n, c=count(): n-next(c)))
l outputs as: 1-2,4-6,3
You can customize the functions depending on your output.

We can use known mathematics formula for checking consecutiveness,
Assuming min number always start from 1
sum of consecutive n numbers 1...n = n * (n+1) /2
def check_is_consecutive(l):
maximum = max(l)
if sum(l) == maximum * (maximum+1) /2 :
return True
return False

Once you verify that the list has no duplicates, just compute the sum of the integers between min(l) and max(l):
def check(l):
total = 0
minimum = float('+inf')
maximum = float('-inf')
seen = set()
for n in l:
if n in seen:
return False
seen.add(n)
if n < minimum:
minimum = n
if n > maximum:
maximum = n
total += n
if 2 * total != maximum * (maximum + 1) - minimum * (minimum - 1):
return False
return True

import numpy as np
import pandas as pd
(sum(np.diff(sorted(l)) == 1) >= n) & (all(pd.Series(l).value_counts() == 1))
We test both conditions, first by finding the iterative difference of the sorted list np.diff(sorted(l)) we can test if there are n consecutive integers. Lastly, we test if the value_counts() are all 1, indicating no repeats.

I split your query into two parts part A "list contains up to n consecutive numbers" this is the first line if len(l) != len(set(l)):
And part b, splits the list into possible shorter lists and checks if they are consecutive.
def example (l, n):
if len(l) != len(set(l)): # part a
return False
for i in range(0, len(l)-n+1): # part b
if l[i:i+3] == sorted(l[i:i+3]):
return True
return False
l = [1, 3, 5, 2, 4, 6]
print example(l, 3)

def solution(A):
counter = [0]*len(A)
limit = len(A)
for element in A:
if not 1 <= element <= limit:
return False
else:
if counter[element-1] != 0:
return False
else:
counter[element-1] = 1
return True

The input to this function is your list.This function returns False if the numbers are repeated.
The below code works even if the list does not start with 1.
def check_is_consecutive(l):
"""
sorts the list and
checks if the elements in the list are consecutive
This function does not handle any exceptions.
returns true if the list contains consecutive numbers, else False
"""
l = list(filter(None,l))
l = sorted(l)
if len(l) > 1:
maximum = l[-1]
minimum = l[0] - 1
if minimum == 0:
if sum(l) == (maximum * (maximum+1) /2):
return True
else:
return False
else:
if sum(l) == (maximum * (maximum+1) /2) - (minimum * (minimum+1) /2) :
return True
else:
return False
else:
return True

1.
l.sort()
2.
for i in range(0,len(l)-1)))
print(all((l[i+1]-l[i]==1)

list must be sorted!
lst = [9,10,11,12,13,14,15,16]
final = True if len( [ True for x in lst[:-1] for y in lst[1:] if x + 1 == y ] ) == len(lst[1:]) else False
i don't know how efficient this is but it should do the trick.

With sorting
In Python 3, I use this simple solution:
def check(lst):
lst = sorted(lst)
if lst:
return lst == list(range(lst[0], lst[-1] + 1))
else:
return True
Note that, after sorting the list, its minimum and maximum come for free as the first (lst[0]) and the last (lst[-1]) elements.
I'm returning True in case the argument is empty, but this decision is arbitrary. Choose whatever fits best your use case.
In this solution, we first sort the argument and then compare it with another list that we know that is consecutive and has no repetitions.
Without sorting
In one of the answers, the OP commented asking if it would be possible to do the same without sorting the list. This is interesting, and this is my solution:
def check(lst):
if lst:
r = range(min(lst), max(lst) + 1) # *r* is our reference
return (
len(lst) == len(r)
and all(map(lst.__contains__, r))
# alternative: all(x in lst for x in r)
# test if every element of the reference *r* is in *lst*
)
else:
return True
In this solution, we build a reference range r that is a consecutive (and thus non-repeating) sequence of ints. With this, our test is simple: first we check that lst has the correct number of elements (not more, which would indicate repetitions, nor less, which indicates gaps) by comparing it with the reference. Then we check that every element in our reference is also in lst (this is what all(map(lst.__contains__, r)) is doing: it iterates over r and tests if all of its elements are in lts).

l = [1, 3, 5, 2, 4, 6]
from itertools import chain
def check_if_consecutive_and_no_duplicates(my_list=None):
return all(
list(
chain.from_iterable(
[
[a + 1 in sorted(my_list) for a in sorted(my_list)[:-1]],
[sorted(my_list)[-2] + 1 in my_list],
[len(my_list) == len(set(my_list))],
]
)
)
)
Add 1 to any number in the list except for the last number(6) and check if the result is in the list. For the last number (6) which is the greatest one, pick the number before it(5) and add 1 and check if the result(6) is in the list.

Here is a really short easy solution without having to use any imports:
range = range(10)
L = [1,3,5,2,4,6]
L = sorted(L, key = lambda L:L)
range[(L[0]):(len(L)+L[0])] == L
>>True
This works for numerical lists of any length and detects duplicates.
Basically, you are creating a range your list could potentially be in, editing that range to match your list's criteria (length, starting value) and making a snapshot comparison. I came up with this for a card game I am coding where I need to detect straights/runs in a hand and it seems to work pretty well.

most efficient way to find a sum of two numbers

I am looking into a problem: given an arbitrary list, in this case it is [9,15,1,4,2,3,6], find any two numbers that would sum to a given result (in this case 10). What would be the most efficient way to do this? My solution is n2 in terms of big O notation and even though I have filtered and sorted the numbers I am sure there is a way to do this more efficiently. Thanks in advance
myList = [9,15,1,4,2,3,6]
myList.sort()
result = 10
myList = filter(lambda x:x < result,myList)
total = 0
for i in myList:
total = total + 1
for j in myList[total:]:
if i + j == result:
print i,j
break

O(n log n) solution
Sort your list. For each number x, binary search for S - x in the list.
O(n) solution
For each number x, see if you have S - x in a hash table. Add x to the hash table.
Note that, if your numbers are really small, the hash table can be a simple array where h[i] = true if i exists in the hash table and false otherwise.

Use a dictionary for this and for each item in list look for total_required - item in the dictionary. I have used collections.Counter here because a set can fail if total_required - item is equal to the current item from the list. Overall complexity is O(N):
>>> from collections import Counter
>>> def find_nums(total, seq):
c = Counter(seq)
for x in seq:
rem = total - x
if rem in c:
if rem == x and c[rem] > 1:
return x, rem
elif rem != x:
return x, rem
...
>>> find_nums(2, [1, 1])
(1, 1)
>>> find_nums(2, [1])
>>> find_nums(24, [9,15,1,4,2,3,6])
(9, 15)
>>> find_nums(9, [9,15,1,4,2,3,6])
(3, 6)

I think, this solution would work....
list = [9,15,1,4,2,3,6]
result = 10
list.sort()
list = filter(lambda x:x < result,list)
myMap = {}
for i in list:
if i in myMap:
print myMap[i], i
break
myMap[result - i] = i

how to find the middle number in python

Given 3 numbers, I need to find which number lies between the two others.
ie,given 3,5,2
I need 3 to be returned.
I tried to implement this by going thru all three and using if else conditions to check if each is between the other two.But this seems a naive way to do this.Is there a better way?

Put them in a list, sort them, pick the middle one.

>>> x = [1,3,2]
>>> sorted(x)[len(x) // 2]
2

The fastest obvious way for three numbers
def mean3(a, b, c):
if a <= b <= c or c <= b <= a:
return b
elif b <= a <= c or c <= a <= b:
return a
else:
return c

What you want is the median. You can use this code below for any number of numbers:
import numpy
numbers = [3,5,2]
median = numpy.median(numbers)
for a custom solution you can visit this page.

You could do
numbers = [3, 5, 2]
sorted(numbers)[1]

This is a O(n) implementation of the median using cumulative distributions. It's faster than sorting, because sorting is O(ln(n)*n).
def median(data):
frequency_distribution = {i:0 for i in data}
for x in data:
frequency_distribution[x] =+ 1
cumulative_sum = 0
for i in data:
cumulative_sum += frequency_distribution[i]
if (cumulative_sum > int(len(data)*0.5)):
return i

Check this (Suppose list already sorted):
def median(list):
ceil_half_len = math.ceil((len(list)-1)/2) # get the ceil middle element's index
floor_half_len = math.floor((len(list)-1)/2) # get the floor middle element 's index'
return (list[ceil_half_len] + list[floor_half_len]) / 2

If it's just for 3 numbers. First find the max and the min numbers in list. Remove them, the remainder of list is the medium. If you don't want to use the "if-else" or "sorted()" that is.
def mid(a,b,c):
num = [a,b,c]
small, big = min(num), max(num)
num.remove(small)
num.remove(big)
return num[0] # return the remainder

Here is my attempt using a more pythonic version
def median(a):
sorted_a = sorted(a)
if len(a) % 2 == 0:
median = sum(sorted_a[(len(a)//2)-1:(len(a)//2)+1])/2.
else:
median = sorted_a[(len(a)-1)//2]
>>> x = [64630, 11735, 14216, 99233, 14470, 4978, 73429, 38120, 51135, 67060]
>>> median(x)
>>> 44627.5
>>> y = [1, 2, 3]
>>> median(y)
>>> 2

If you wish to avoid sorting, you can do:
def find_median(x):
return sum(x) - max(x) - min(x)