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I have run the the following functions and studied them line-by-line. I understand how the outer for-loop in f(n) works and how the while-loop works in g(n) but I understand the role of the inner for-loop in f(n). Also, how do these loops work with t = t+1? Thanks in advance!
def f(n):
t=0
for i in range(n):
for j in range(2*i):
t=t+1
return t
f(5)
def g(n):
t=0
j=n
while j>1:
t = t+1
j = j/2
return t
g(32)
The inner loop keeps adding 1 to t until it adds 2 times each item from the outer loop. So it adds 0 + 2 + 4 + 6 + 8. The range(5) is similar to a list equivalent to [0,1,2,3,4].
t=t+1 simply adds 1 to the value of t every time the line is run.
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Video Puzzle
Current code:
for i in range(6,0,-2):
Spaceship.step(2)
Dev.step(i)
for idk in range(3):
Dev.turnRight()
Dev.step(i*2)
Dev.turnRight()
Dev.step(i)
In this puzzle the objective is to get all the item (blue thing). With 6 line of code, and I'm currently at 8 line of code. I don't know how to minimalize the line of code.
Note:
Dev.step() is the robot, it can go backward by set the value by negative.
Spaceship.step() is the spaceship, it can not go backward.
You can avoid pythonesque code like so:
for i in range(6,0,-2):
Spaceship.step(2)
for idk in range(4):
Dev.step(i)
Dev.turnRight()
Dev.step(i)
This is a possible solution in only 6 lines:
for i in range(6, 0, -2):
Spaceship.step(2)
for k, j in enumerate([1, 2, 2, 2, 1]):
Dev.step(i * j)
if k != 4:
Dev.turnRight()
The idea is to group all steps of the robot in a list in order to do a nested loop and turn only if it is not the last element of the list.
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Example:
import numpy
a = [1,2,3,4,5]
b = []
for i in range(len(a)):
b.append(a[i]+1)
Here, I have b = [2,3,4,5,6].
But I want sum multi times...(maybe I want to sum with 3 times). I expected after three times sum I have b = [4,5,6,7,8].
So, how I can get b=[4,5,6,7,8] from a=[1,2,3,4,5] with 3 times add 1 with loop?
Add 3 to each element using a list comprehension;
b = [i+3 for i in a]
or as a function, where you can change the value added to the list easily;
def add_k_to_list(k, a_list):
return [i+k for i in a_list]
To repeatedly apply as per your comment;
def add_k_to_list(k, a_list):
for _ in range(k):
a_list = [i+1 for i in a_list]
return(a_list)
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firstLst = [1,2,3,4]
secondLst = [5,6,3,2,4,1]
If i am passing the two lists above as arguements in a function. I would like to know how I can check if all elements in the firstLst are in the secondLst using recursions with Python. I can achieve this by having two extra parameters i and j which will allow me to check each element in firstLst and secondLst, but is there a way to do this by only having the two lists as parameters?
example of the function with parameters:
def firstInSecond(lst1,lst2):
How can True be returned if all elements in lst1 are in lst2 otherwise False is returned. I hope this makes sense.
This isn't exactly recursive, but simpler.
set(firstLst).issubset(set(secondLst))
For a recursive answer,
def firstInSecond(lst1,lst2):
if len(lst1) == 0:
return True
if lst1[0] in lst2:
return firstInSecond(lst1[1:], lst2)
else:
return False
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if I have
list=[1,2,3,4,5,6]
How can i make it
list=123456
Thanks for your help in advance!
You can do this:
inlist=[1,2,3,4,5,6]
length = len(inlist)
s = 0
for i in range(length):
s += (inlist[i] * ( 10 ** (length-1-i)))
inlist = s
print(inlist)
This will give you:
123456
You need to utilize the power of 10 to multiply it with each number.
Note that you shouldn't use list as a variable name as it is a Python keyword.
Another version (without using any built-in functions at all):
inlist=[1,2,3,4,5,6]
count = 1
s = 0
for elem in inlist[::-1]:
s += (elem * ( 10 ** (count-1)))
count += 1
inlist = s
print(inlist)
you can do it by for and join likes the following:
int(''.join([str(i) for i in my_list]))
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Write a function called sumDivs that accepts a single, positive integer and calculates the
sum of all of its proper divisors. Use a for loop to find and sum the proper divisors of the
given integer.
For example,
>>> sumDivs(8)
7
this is what i have:
def sums(n):
i=0
for i in range (1, n-1):
if n % i == 0:
return n-1
i+=1
You shouldn't return the first divisor you find, you should be adding them as you go; i takes on the values from 1 to n-2 in your loop change to for i in range(1,n); and there is no need to define an initial value for i. Try this instead:
def sums(n):
sum=0
for i in range (1, n):
if n % i == 0:
sum+=i
return sum