I'm still a beginner so please bear with me. So I'm trying to get a list of factorial numbers from 1 to 5.
factorial=[]
for i in range(1,5):
for x in range(1,5):
while(i>x):
factorial.append(i*x)
When I switch out the factorial.append for print it just continously spits out 2s, does anyone know why, and if so what to do to fix this and what other viable method is there to get a list of factorial numbers?
I this case, I recommend you to use a recursive function:
def factorial(x):
if x == 1:
return 1
else:
return x*factorial(x-1)
Ex:
>>>factorial(5)
120
>>>factorial(10)
3628800
you can do something like :
>>> f=[1] # initialized your list
>>> for i in range(5): # take each value in the list [0,1,2,3,4]
... f.append(f[i]*(i+1)) #to create a next point multiply with the last value
...
>>> f=f[1:] #don't keep the first (repeated) point
>>> f #you have your list !
[1, 2, 6, 24, 120]
You're getting stuck in your while loop.
If you go step by step through your code, you'll see what is going on:
In the first loop, i value will be 1, and x values from 1 to 5, so i will never be > than x, thus you won't enter the while loop.
At the start of the second loop, i value will be 2 and x value will be 1, so you will enter the while loop.
You will stay in stay in the while loop until i becomes lower or equal to x, but this will never happen because to continue with the for loop you'll need to exit the while loop first.
Despite the error, i do not understand the reasoning that brought you there.
The common way to handle factorials is with recursion, as #Julio CamPlaz's answer shows, but it may simplier to understand the following:
# in each for loop, you multiply your n variable for the i value
# and you store the result of the moltiplication in your n variable
n=1
for i in range(1,6): # if you want the numbers from 1 to 5, you have to increase by one your maximum value
n = n*i
print n
Related
I have been working on this assignment for about 2 weeks and have nothing done. I am a starter at coding and my teacher is really not helping me with it. She redirects me to her videos that I have to learn from every time and will not directly tell or help me on how I can do it. Here are the instructions to the assignment (said in a video, but made it into text.
Find the mean
Create a program that finds the mean of a list of numbers.
Iterate through it, and instead of printing each item, you want to add them together.
Create a new variable inside of that, that takes the grand total when you add things together,
And then you have to divide it by the length of your array, for python/java script you’ll need to use the method that lets you know the length of your list.
Bonus point for kids who can find the median, to do that you need to sort your list and then you need to remove items from the right and the left until you only have one left
All you’re doing is you need to create a variable that is your list
Create another variable that is a empty one at the moment and be a number
Iterate through your list and add each of the numbers to the variable you created
Then divide the number by the number of items that you had in the list.
Here's what I've done so far.
num = [1, 2, 3, 4, 5, 6];
total = 0;
total = (num[0] + total)
total = (num[1] + total)
total = (num[2] + total)
total = (num[3] + total)
total = (num[4] + total)
total = (num[5] + total)
print(total)
However, she tells me I need to shorten down the total = (num[_] + total) parts into a loop. Here is how she is telling me to do a loop in one of her videos.
for x in ____: print(x)
or
for x in range(len(___)): print (x+1, ____[x])
there is also
while i < len(___):
print(___[i])
i = i + 1
I really don't understand how to do this so explain it like I'm a total noob.
First of all, loops in python are of two types.
while: a while loop executes the code in a body until a condition is true. For example:
i = 0
while(i < 5):
i = i + 1
executes i = i + 1 until i < 5 is true, meaning that when i will be equal to 5 the loop will terminate because its condition becomes false.
for: a for loop in python iterates over the items of any sequence, from the first to the last, and execute its body at each iteration.
Note: in both cases, by loop body I mean the indented code, in the example above the body is i = i + 5.
Iterating over a list. You can iterate over a list:
Using an index
As each position of the array is indexed with a positive number from 0 to the length of the array minus 1, you can access the positions of the array with an incremental index. So, for example:
i = 0
while i < len(arr):
print(arr[i])
i = i + 1
will access arr[0] in the first iteration, arr[1] in the second iteration and so on, up to arr[len(arr)-1] in the last iteration. Then, when i is further incremented, i = len(arr) and so the condition in the while loop (i < arr[i]) becomes false. So the loop is broken.
Using an iterator
I won't go in the details of how an iterator works under the surface since it may be too much to absorb for a beginner. However, what matters to you is the following. In Python you can use an iterator to write the condition of a for loop, as your teacher showed you in the example:
for x in arr:
print(x)
An iterator is intuitively an object that iterates over something that has the characteristic of being "iterable". Lists are not the only iterable elements in python, however they are probably the most important to know. Using an iterator on a list allows you to access in order all the elements of the list. The value of the element of the list is stored in the variable x at each iteration. Therefore:
iter 1: x = arr[0]
iter 2: x = arr[1]
...
iter len(arr)-1: x = arr[len(arr)-1]
Once all the elements of the list are accessed, the loop terminates.
Note: in python, the function range(n) creates an "iterable" from 0 to n-1, so the for loop
for i in range(len(arr)):
print(arr[i])
uses an iterator to create the sequence of values stored in i and then i is in turn used on the array arr to access its elements positionally.
Summing the elements. If you understand what I explained to you, it should be straightforward to write a loop to sum all the elements of a list. You initialize a variable sum=0 before the loop. Then, you add the element accessed as we saw above at each iteration to the variable sum. It will be something like:
sum = 0
for x in arr:
sum = sum + x
I will let you write an equivalent code with the other two methods I showed you and do the other points of the assignment by yourself. I am sure that once you'll understand how it works you'll be fine. I hope to have answered your question.
She wants you to loop through the list.
Python is really nice makes this easier than other languages.
I have an example below that is close to what you need but I do not want to do your homework for you.
listName = [4,8,4,7,84]
for currentListValue in listName:
#Do your calculating here...
#Example: tempVar = tempVar + (currentListValue * 2)
as mentioned in the comments w3schools is a good reference for python.
lst = [3, 4, 1, 2, 9]
givenSum = 12
table = {}
x = 0
y = 0
for i in range(0, len(lst)):
table[givenSum - lst[i]] = 1
i += 1
for x in table:
for y in table:
if (x + y) == givenSum:
print(x, "and", y, "is equal to", givenSum)
break
This is the output
9 and 3 is equal to 12
3 and 9 is equal to 12
I don't know why it's being shown up twice. I need to find a pair of values that add up to the given sum and this is the only way I could think of. I only want it to show up once though any ideas on how I can do that?
There are better solutions, but to fix your issue making minimal changes to your code:
lst = [3, 4, 1, 2, 9]
givenSum = 12
for x in range(0, len(lst) - 1):
for y in range(x + 1, len(lst)):
if lst[x] + lst[y] == givenSum:
print(lst[x], "and", lst[y], "is equal to", givenSum)
break
This will print
3 and 9 is equal to 12
Note that the redundant table is completely removed from the code.
If you run it for a better test case:
lst = [3, 4, 5, 6, 7, 1, 2, 9]
it will print
3 and 9 is equal to 12
5 and 7 is equal to 12
First, to address why the looping continues and gives a second output, break can only break out of its immediate loop. Since you have a nested loop, the break only stops the for y in table: inner loop, but allows for x in table outer loop to move onto it's next iteration. So eventually, x is able to take the value of 3 later on, thus giving you the two outputs you see.
So, if you need a way to stop the iteration entirely when a solution is found, you need to either chain the break statements using a for else syntax (which arguably might be tough to read) as follows,
for x in table:
for y in table:
if (x + y) == givenSum:
print(x, "and", y, "is equal to", givenSum)
break #breaks from inner loop
else: #for else syntax: this block runs if and only if there was no break encountered during looping.
continue #jumps the outer loop to next iteration
break #this break is set at outer loop's level. Essentially, we can only reach this portion if there is a break in the inner loop.
For else says: run through the whole iteration, and if no break is found, executes the code in the else block. Essentially, the "else" of a "for else" is like a "for - no break".
However, one easier alternative is to use a function with a return (which also makes it easier to read the code).
def find_given_sum(lst, givenSum):
table = {}
x = 0
y = 0
for i in range(0, len(lst)):
table[givenSum - lst[i]] = 1
i += 1
for x in table:
for y in table:
if (x + y) == givenSum:
print(x, "and", y, "is equal to", givenSum)
return #this returns immediately out of the function, thus stopping the iteration.
Also, you could just repeat the break condition, but repeating code is generally not a good practice.
Hope this helps address why the two outputs are being printed. Now, as for the solution itself, there's actually a much better way to solve this. It builds upon the idea of compliments, which you seem to have a sense of in your table. But it doesn't require iteration over the table itself. As a hint: the ideal solution runs in O(n) time. I will not discuss the ideal solution, but hope this prompts you to find the best approach.
Looping twice for n elements costs you O(N^2) time, which is inefficient for large lists. Modified and tested the code to use hash map/dictionary to store the list elements and now it will take only O(N) time.
Map = dict()
lst = [3, 4, 1, 2, 9]
givenSum = 12
for i in range(0, len(lst)):
rem=givenSum-lst[i]
if rem in Map:
print lst[i],lst[Map[rem]]
else:
Map[lst[i]]=i
Store the value of each list element into the map whenever it does not exist in the map.
At each iteration, take the difference between givenSum and current element at that iteration and then search for that difference in the Map.
If it exists it means the pair exists or else not.
In this approach you are running the loop only once, which takes O(N) time because accessing elements in hash map is O(1)/constant time.
Use itertools to get the result
import itertools
sum = 10
lst = [3, 4, 1, 2, 9]
ans = list(filter(lambda x : x[0]+x[1] == sum,itertools.combinations(lst, 2)))
That's my first question, I'm new at programming, sorry about any inconvinience !
I need to finish an exercise that consists in create a fuction to find the higher number and another function to find the lower number on a List, but when I print the results, it keeps giving me a wrong answer.
Here is my code:
lista_numeros = [1593, 200, 72, 8, 423, 2, 39, 293, 20120]
def menor_numero(arg1):
colector1 = arg1[0]
print(coletor1) # - Checking value added to COLECTOR1
for a in range(len(arg1)):
print(arg1[a]) # - Checking if the A argument is running through the list.
if colector1 < arg1[a]:
colector1 = arg1[a]
return colector1
resultado2 = menor_numero(lista_numeros)
print("menor ", str(resultado2)) # Result is the last position of the list, must be the 6th position.
Thank you very much.
Fist of all indentation is very important in python to tell it the order to execute your code and also to define where code sits within a loop etc.
Now you say you want to make a function that finds the smallest and largest number from the output of another function, for this I will assume this output is a list.
Please see code below with comments.
Mylist = [1, 2, 3, 4, 5, 6, 7, 8, 9] #assume this is output from other funtion
def func(alist): #define function
collector1 = 100 #keeping your collector idea
for i in alist: #iterate through input
if i < collector1: #check if the item you are currently looking at is smaller than the item currently stored in collector
collector1 = i #if is smaller overwitre colletor with new item
print(collector1) #after iterating through all items in input print final value of colletor
func(Mylist) #call function with input
This outputs,
1
Simply change this,
if i > collector1:
And it will now find the largest in input, so output is now.
9
Edit: if you are looking for the smallest number start collector1 at a large number, if you are looking for the largest start collector1 at = 1.
assuming your input is a list and not a string or some sort you can just use the min()/max() methods:
myList = [1,2,3,4,5]
print(max(myList)) >> 5
print(min(myList)) >> 1
you can find more info here :
https://www.tutorialspoint.com/python3/list_max.htm
https://www.tutorialspoint.com/python3/list_min.htm
Your function is finding the maximum instead of the minimum value. Changing the < to > should do what you want.
Python also has builtin methods min() and max() which should do what you want.
#! python3
import random
numbers = []
max = 1000
min = 0
for i in range(40):
numbers.append(random.randint(min,max))
maxNum = min
minNum = max
for num in numbers:
if num > maxNum:
maxNum = num
elif num < minNum:
minNum = num
print(maxNum, minNum)
This is my code, I use the random library in python to generate a random list of numbers, then I set max to equal the maximum number in that list.
The following for loop generates 40 random numbers and appends them to my list.
I then set maxNum to equal zero, so everything will be greater than it, thereby the initial value will not influence the outcome, and then I set minNum to equal max, so every number will be less than it.
The last block of code loops through the numbers list, and compares every number with the current maxNum and minNum variables to see whether that number is greater than the max or less than the min. If it is, the maxNum (or minNum) number will be updated and the code will move on to the next number.
The last statement prints or minimum and maximum values.
I don't know what course you are taking, but I would advise you to familiarize yourself with this code and understand what it is doing, as it's pretty basic and the stuff you encounter in the future will be much harder.
I made a program and I might be making a conceptual error can anyone help me to figure it out?
l=[1,7,6,2,3]
x=0
y=0
for i in l:
if(i>y):
x=i
else:
x=y
y=i
print (x)
The output it gives is 3 i.e. the last number in the list
CAN ANYONE POINT OUT MY MISTAKE PLSS
This code will work for both positive and negative entries.
l=[1,7,6,2,3]
MAX = float('-inf')
for i in l:
if(MAX < i):
MAX = i
print(MAX)
Output: 7
You can also solve this without a for-loop.
like this:
Code
l=[1,7,6,2,3]
print(max(l))
Output
7
Your mistake was
y = i
in line 13, you should do
y = x
instead.
However, if you do not need to use loop for some bizarre reason, the best solution would be to use max function, as others already pointed out.
If you prefer to use your function, I would optimise it even further, for example:
l = (-1000, 1, 7, 6, 2, 3, 1000)
x = float('-inf')
for i in l:
if i > x:
x = i
print x
The following loop will work,
l=[1,7,6,2,3]
x=0
for i in l:
if(i>x):
x=i
print (x)
You can also use the max() method to find the largest element in a list,
`max(l)` will return `7`.
l=[1,7,6,2,3]
max=0 # set the max of the list to 0 to start
for i in l: # run through the entire list
if(max<i): # check to see if the present element > max
max=i # if so, set max to that value
print (max)
I want to know if is it possible to change the value of the iterator in its for-loop?
For example I want to write a program to calculate prime factor of a number in the below way :
def primeFactors(number):
for i in range(2,number+1):
if (number%i==0)
print(i,end=',')
number=number/i
i=i-1 #to check that factor again!
My question : Is it possible to change the last two line in a way that when I change i and number in the if block, their value change in the for loop!
Update: Defining the iterator as a global variable, could help me? Why?
Short answer (like Daniel Roseman's): No
Long answer: No, but this does what you want:
def redo_range(start, end):
while start < end:
start += 1
redo = (yield start)
if redo:
start -= 2
redone_5 = False
r = redo_range(2, 10)
for i in r:
print(i)
if i == 5 and not redone_5:
r.send(True)
redone_5 = True
Output:
3
4
5
5
6
7
8
9
10
As you can see, 5 gets repeated. It used a generator function which allows the last value of the index variable to be repeated. There are simpler methods (while loops, list of values to check, etc.) but this one matches you code the closest.
No.
Python's for loop is like other languages' foreach loops. Your i variable is not a counter, it is the value of each element in a list, in this case the list of numbers between 2 and number+1. Even if you changed the value, that would not change what was the next element in that list.
The standard way of dealing with this is to completely exhaust the divisions by i in the body of the for loop itself:
def primeFactors(number):
for i in range(2,number+1):
while number % i == 0:
print(i, end=',')
number /= i
It's slightly more efficient to do the division and remainder in one step:
def primeFactors(number):
for i in range(2, number+1):
while True:
q, r = divmod(number, i)
if r != 0:
break
print(i, end=',')
number = q
The only way to change the next value yielded is to somehow tell the iterable what the next value to yield should be. With a lot of standard iterables, this isn't possible. however, you can do it with a specially coded generator:
def crazy_iter(iterable):
iterable = iter(iterable)
for item in iterable:
sent = yield item
if sent is not None:
yield None # Return value of `iterable.send(...)`
yield sent
num = 10
iterable = crazy_iter(range(2, 11))
for i in iterable:
if not num%i:
print i
num /= i
if i > 2:
iterable.send(i-1)
I would definitely not argue that this is easier to read than the equivalent while loop, but it does demonstrate sending stuff to a generator which may gain your team points at your next local programming trivia night.
It is not possible the way you are doing it. The for loop variable can be changed inside each loop iteration, like this:
for a in range (1, 6):
print a
a = a + 1
print a
print
The resulting output is:
1
2
2
3
3
4
4
5
5
6
It does get modified inside each for loop iteration.
The reason for the behavior displayed by Python's for loop is that, at the beginning of each iteration, the for loop variable is assinged the next unused value from the specified iterator. Therefore, whatever changes you make to the for loop variable get effectively destroyed at the beginning of each iteration.
To achieve what I think you may be needing, you should probably use a while loop, providing your own counter variable, your own increment code and any special case modifications for it you may need inside your loop. Example:
a = 1
while a <= 5:
print a
if a == 3:
a = a + 1
a = a + 1
print a
print
The resulting output is:
1
2
2
3
3
5
5
6
Yes, we can only if we dont change the reference of the object that we are using. If we can edit the number by accessing the reference of number variable, then what you asked is possible.
A simple example:
a=[1,2,3]
a=a+[4]==>here, a new object is created which plots to different address.
a+=[4]==>here , the same object is getting updated which give us the desired result.
number=10
list1=list(range(2,number+1))
# list1
for i in list1:
print(list1,i)
if (number%i==0):
print(i,end=',')
number=number//i #we can simply replace it with number//=i to edit the number without changing the reference or without creating a new object.
try:
[list1.pop() for i in range(10,0,-1) if(i>number)]
#here pop() method is working on the same object which list created by number refers. so, we can able to change the iterable in the forloop.
except:
continue
i=i-1 #to check that factor again!