Iterative summation - python

I'm trying to write a python code that allows me to iteratively sum up the average values of three elements of a list, starting with the third element and its two predecessors. Let me give you an example:
list = [1, 2, 3, 4, 5, 6, 7]
I want to calculate the following:
sum_of_average_values = sum(1, 2, 3)/3 + sum(2, 3, 4)/3 + sum(3, 4, 5)/3 + sum(4, 5, 6)/3 + sum(5, 6, 7)/3
Since I'm quite new to programming I couldn't find an effective way of putting this into a function.

You can do in this way:
a = [1,2,3,4,5,6,7]
sum_of_average_values = 0
for i in range(0,len(a)-2):
sum_of_average_values += sum(a[i:i+2])/3
print(sum_of_average_values)

You could use rolling from pandas.
import pandas as pd
num_list = [1, 2, 3, 4, 5, 6, 7]
average_sum = sum(pd.Series(num_list).rolling(3).mean().dropna())
print(average_sum)

Many ways to achieve this, one way is recursively.
The function averages the last three elements of a list and adds the result to the result generated by the function with a list lacking the last element. Continues like this until the list is shorter than 3.
def fn(l):
if len(l) < 3:
return 0
return sum(l[-3:])/3 + fn(l[:-1])
print(fn([1, 2, 3, 4, 5, 6, 7]))

Another solution where you can specify the amount of elements you want to sum up and average:
l = [1, 2, 3, 4, 5, 6, 7]
def sum_avg(l, n):
res = 0
for i in range(n-1, len(l)):
res += sum([l[j] for j in range(i, i-n, -1)])/n
return res
print(sum_avg(l, 3))
--> 20.0

Mathematically, this would could be obtain by averaging the sums of 3 sublists:
L = [1, 2, 3, 4, 5, 6, 7]
r = (sum(L) + sum(L[1:-1]) + sum(L[2:-2]))/3 # 20.0
and can be generalized to a window size of w:
w = 3
r = sum(sum(L[p:-p or None]) for p in range(w)) / w
It can also be implemented without the overhead of generating sublists by using item positions to determine the number of times they are added to the total:
r = sum(n*min(i+1,len(L)-i,w) for i,n in enumerate(L)) / w
This would be the most memory-efficient of the 3 methods because it use an iterator to feed data to the sum function and only goes through the data once.
Detailed explanation:
Since all the averages that are added together are a division by 3, we can produce the total sum and divide by 3 at the end
the number at the first and last positions are added once
the number at the second and penultimate positions are added twice
The numbers from the third position up to the antepenultimate will be added 3 times
visually:
(1 + 2 + 3) / 3
(2 + 3 + 4) / 3
(3 + 4 + 5) / 3
(4 + 5 + 6) / 3
(5 + 6 + 7) / 3
(1x1 + 2x2 + 3x3 + 4x3 + 5x3 + 6x2 + 7x1) / 3 = 20.0
n = 1 2 3 4 5 6 7 # value
* = 1 2 3 3 3 2 1 # multiplier (times added)
-------------------------
(2, 4, 9, 12, 15, 12, 7) / 3 = 20.0
i = 0 1 2 3 4 5 6 # index
1 2 3 3 3 2 1 # min(i+1,len(L)-i,w) = multiplier

You can do in one line using list comprehension as:
n = 3
avg = sum( [ sum(lst[i:i+n])/n for i in range(0, len(lst) - (n - 1)) ] )
print(avg) # 20.0

Related

Creating code to make sum of array elements with variable

I want to create a code that has the variable how many elements you take from an list and sum those. For example, if I type in 3, it takes the first 3 elements of an array and addes them together.
Preferably in some for loop but who knows what kind of creative solutions I get :)
x = [1, 3, 4, 2, 6, 9, 4]
Amount = 3
Sum = 0
Sum += x[Amount-3] + x[Amount-2] + x[Amount-1]
Desired result: 8
Amount_2 = 4
Sum = 0
Sum += x[Amount-4] + x[Amount-3] + x[Amount-2] + x[Amount-1]
Desired result: 11
Hope this explains it well.
x = [2, 6, 7, 7, 12] # your list here
amount = int(input("Enter amount: "))
print(f'desired result: {sum(x[:amount])}')

How to make with lists?

How to make a multiplication chart with nested lists and for ? I need to all numbers from first list multiply to from second list
chart = [
[],
[],
]
for i in range(1,len(chart)+1):
for j in range(i,i*len(chart)+1):
print(f'{i} * {j} = {i*j}')
In python positions of elements in a list start from 0.
The chart list contains 2 lists:
chart[0] = [1, 2, 3, 4, 5]
chart[1] = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
What you want to do is access the elements of the first list and multiply by elements of the second list.
for i in range(len(chart[0])): # range(5) => 0, 1, 2, 3, 4
for j in range(len(chart[1])): # range(10) => 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
print(f'{chart[0][i]} * {chart[1][j]} = {chart[0][i] * chart[1][j]}
itertools.product will produce the desired output, creating 2-tuples consisting of one element from the first list and one element from the second list:
import itertools
chart = [
[1,2,3,4,5],
[1,2,3,4,5,6,7,8,9,10],
]
for i, j in itertools.product(*chart):
print(f'{i} * {j} = {i*j}')
Use a cartesian product:
chart = [[1,2,3,4,5],[1,2,3,4,5,6,7,8,9,10]]
>>> print('\n'.join([f'{i} * {j} = {i*j}' for i in chart[0] for j in chart[1]]))
1 * 1 = 1
1 * 2 = 2
1 * 3 = 3
1 * 4 = 4
...
4 * 10 = 40
5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Find 3 maximum elements of list with python

How can I search for the three maximum elements of a list and replace it at same index with its result when divided by 2.
Please what am I doing wrong:
input is: 2 5 8 19 1 15 7 20 11
output should be : 2 5 8 9.5 1 7.5 7 10 11
Index out of range is the result displayed
def numberInput(line):
lineInput = [int(i) for i in line.split()]
min1 = min(lineInput)
for j in range(len(lineInput)):
if lineInput[j]>min1 and lineInput[j] > lineInput[j+1]:
max1 = lineInput[j]/float(2)
else:
max1 = lineInput[j]/float(2)
lineInput[j] = max1
lineInput[j] = max1
return(lineInput)
number = '2 5 8 19 1 15 7 20 11'
print(numberInput(number))
If the order of list isn't important, you can simply sort the list in descending order and then replace the first 3 elements as
a = [2, 5, 8, 19, 1, 15, 7, 20, 11]
a.sort(reverse = True)
a[0] = a[0] / 2
a[1] = a[1] / 2
a[2] = a[2] / 2
Output
[10.0, 9.5, 7.5, 11, 8, 7, 5, 2, 1]
If the order is important,
import heapq
largest = heapq.nlargest(3, a)
for i in range(len(a)):
if a[i] in largest:
a[i] = a[i] / 2
Output
[2, 5, 8, 9.5, 1, 7.5, 7, 10.0, 11]
heapq.nlargest() is a function of heapq which can give you n largest numbers from a list. Since lists in Python do not have a replace function, the list had to be traversed once, and then replaced manually. Hope this resolves your issue.
Why wouldn't you just walk through the list once (it maybe a bit longer code, but definitely efficient)
def numberInput(line):
lineInput = [int(i) for i in line.split()]
n = len(lineInput)
if n <= 3:
return [i / 2 for i in lineInput]
max_pairs = [(lineInput[i], i) for i in range(3)] # keep track of 3 max's and their indices
max_pairs.sort(key = lambda x: -x) # sort in descending order
for i in range(3, n):
if lineInput[i] >= max_pairs[0][0]: # greater than the largest element
max_pairs = [(lineInput[i], i)] + max_pairs[:2]
elif lineInput[i] >= max_pairs[1][0]: # greater than second element
max_pairs = [max_pairs[0], (lineInput[i], i), max_pairs[1]]
elif lineInput[i] >= max_pairs[2][0]: # greater than third element
max_pairs = max_pairs[:2] + [(lineInput[i], i)]
for pair in max_pairs:
lineInput[pair[1]] = lineInput[pair[0]] / 2
return lineInput
Explanation: max_pairs is a set of three tuples, containing the maximum three elements and their indices
Note: I think the above is easiest to understand, but you can do it in a loop if you don't like all those ifs

Transforming an array of integers and computing the sum

Suppose we need to transform an array of integers and then compute the sum.
The transformation is the following:
For each integer in the array, subtract the first subsequent integer that is equal or less than its value.
For example, the array:
[6, 1, 3, 4, 6, 2]
becomes
[5, 1, 1, 2, 4, 2]
because
6 > 1 so 6 - 1 = 5
nothing <= to 1 so 1 remains 1
3 > 2 so 3 - 2 = 1
4 > 2 so 4 - 2 = 2
6 > 2 so 6 - 2 = 4
nothing <= to 2 so 2 remains 2
so we sum [5, 1, 1, 2, 4, 2] = 15
I already have the answer below but apparently there is a more optimal method. My answer runs in quadratic time complexity (nested for loop) and I can't figure out how to optimize it.
prices = [6, 1, 3, 4, 6, 2]
results = []
counter = 0
num_prices = len(prices)
for each_item in prices:
flag = True
counter += 1
for each_num in range(counter, num_prices):
if each_item >= prices[each_num] and flag == True:
cost = each_item - prices[each_num]
results.append(cost)
flag = False
if flag == True:
results.append(each_item)
print(sum(results))
Can someone figure out how to answer this question faster than quadratic time complexity? I'm pretty sure this can be done only using 1 for loop but I don't know the data structure to use.
EDIT:
I might be mistaken... I just realized I could have added a break statement after flag = False and that would have saved me from a few unnecessary iterations. I took this question on a quiz and half the test cases said there was a more optimal method. They could have been referring to the break statement so maybe there isn't a faster method than using nested for loop
You can use a stack (implemented using a Python list). The algorithm is linear since each element is compared at most twice (one time with the next element, one time with the next number smaller or equals to it).
def adjusted_total(prices):
stack = []
total_substract = i = 0
n = len(prices)
while i < n:
if not stack or stack[-1] < prices[i]:
stack.append(prices[i])
i += 1
else:
stack.pop()
total_substract += prices[i]
return sum(prices) - total_substract
print(adjusted_total([6, 1, 3, 4, 6, 2]))
Output:
15
a simple way to do it with lists, albeit still quadratic..
p = [6, 1, 3, 4, 6, 2]
out= []
for i,val in zip(range(len(p)),p):
try:
out.append(val - p[[x <= val for x in p[i+1:]].index(True)+(i+1)])
except:
out.append(val)
sum(out) # equals 15
NUMPY APPROACH - honestly don't have alot of programming background so I'm not sure if its linear or not (depending on how the conditional masking works in the background) but still interesting
p = np.array([6, 1, 3, 4, 6, 2])
out = np.array([])
for i,val in zip(range(len(p)),p):
pp = p[i+1:]
try:
new = val - pp[pp<=val][0]
out = np.append(out,new)
except:
out = np.append(out,p[i])
out.sum() #equals 15

How to find the factors of a number and return it in a list in python [duplicate]

This question already has answers here:
What is the most efficient way of finding all the factors of a number in Python?
(29 answers)
Closed 9 years ago.
This is the code I have right now. I can't get it to return the right results for the question.
def problem(n):
myList = [1,n]
for i in range(1,n):
result = int(n ** .5)
new = n/result
i = i + 1
myList.append(new)
return myList
Factors of n are all numbers that divide into n evenly. So i is a factor of n if n % i == 0.
You need to do is perform this test for each number from 1 to n, and if that condition is true append that number to your list.
If you have issues as you start to write this code, update your question with what you tried.
Note that the above approach is not the most efficient way to find factors, but it seems to me like this is just an exercise for a beginning programmer so a naive approach is expected.
There are a few problems with your code. First of all you do not need to increment i as your for loop already does that. Secondly, using some basic math principles you only need to go through a range of numbers up to the square root of your passed in number. I will leave the second part for you to play and experiment with.
def problem(n):
myList = []
for i in range(1, n+1):
if n % i == 0:
myList.append(i)
return myList
For a more advanced approach you can try list comprehensions which are very powerful but are usually better for smaller data sets.
def problem(n):
return [x for x in range(1, n+1) if n % x == 0]
You only need to iterate from 1 to n ** 0.5 + 1, and your factors will be all i's, and n/i's you pick up along the way.
For example: factors of 10:
We only need to iterate from 1 to 4
i = 1 => 10 % 1 == 0, so factors: i = 1, 10 / i = 10
i = 2 => 10 % 2 == 0, so factors: i = 2, 10 / i = 5
i = 3 => 10 % 3 != 0, no factors
We don't need to go any further, the answer is 1, 2, 5, 10.
def problem(n):
myList = []
for i in xrange(1, int(n ** 0.5 + 1)):
if n % i == 0:
if (i != n/i):
myList.append(i)
myList.append(n / i)
else:
myList.append(i)
return myList
Result:
>>> problem(10)
[1, 10, 2, 5]
>>> problem(12)
[1, 12, 2, 6, 3, 4]
>>> problem(77)
[1, 77, 7, 11]
>>> problem(4)
[1, 4, 2]
>>> problem(64)
[1, 64, 2, 32, 4, 16, 8]
>>> len(problem(10 ** 12))
169
use a list comprehension:
In [4]: num=120
In [5]: [x for x in range(2,int(num/2)+1) if num%x==0]
Out[5]: [2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60]
In [6]: num=121
In [7]: [x for x in range(2,int(num/2)+1) if num%x==0]
Out[7]: [11]

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