I'm having an issue in python3. I want to exclude 0 as one of the values, however I can't seem to make that happen. I attempted adding another "and if" parameter but the result still included 0.
I want the values that are multiples of 6 between 1 and 100. 0 is clearly not one of them.
x for x in range(100) if x % 6 == 0
The most direct answer is just
range(6, 101, 6)
But your original approach should work if you do something like:
[x for x in range(101) if x % 6 == 0 and x != 0]
(no and if needed, it's just a single if clause with a compound test).
Try:
x for x in range(100) if x%6 == 0 and x != 0
or simply:
x for x in range(1,100) if x%6 == 0
If you want to exclude 0, don't include it in the first place :
range(1, 100)
Related
I want to check the numbers in a list are even or odd in order and if not tell me which one.
for example :
1,2,3,4,5 IS OK
but:
1,2,8,3,4,5 number 8 IS NOT OK
I tried to make the list into 2 different lists and then check if the are all odd or all even but I can't
figure out how to check both of them.
lst = [1,2,3,4,5,6,7,8,9,10]
m = lst[::2]
w = lst[1::2]
for i, j in(m,w):
if i % 2 == 0 and j % 2 == 0:
Try with
list(map(lambda el: 'odd' if el%2 else 'even', lst))
I'm new to programming, and I don't understand why this code returns the binary number of the input.
x = int(input())
while x > 0:
print(x % 2, end='')
x = x // 2
For example, when I enter 6, I expect it to return 0 (6 % 2 = 0, 0 // 2 = 0), but it returns 011.
I stated contemplating this because, in an assignment, we're supposed to use this coding to help us reverse the output (while using a string) but I don't know how to reverse it when I don't know why it works.
The answer of 011 is correct for your code, just trace it thru. You start with 6%2 so you get a 0. Then 6 floor 2 is 3, which is then the x for the next iteration so it becomes 3%2 which is 1 and the final iteration creates the last 1. Remember that you're not changing the value of x in the print statement, it only changes in the last line.
I am doing an online code challenge. I have an array which I need to sort and record to minimum number of iterations required to be sorted. I have the following code.
def minSwap(ar):
c = 0
for i in range(0, len(ar)):
if ar[i] == i+1:
continue
else:
for k in range(i+1, len(ar)):
if ar[k] == i+1:
ar[k] = ar[i]
ar[i] = i+1
c = c+1
break
return c
This code passes majority of test cases, however for really huge number of case such as where array size is beyond (let's say 50000) it gets timeout.
How can I identify the faulty block of code? I can't see a way to tweak it further.
Looking at the problem statement, it looks like you want to sort a list that has numbers starting from 1 thru n.
If you are trying to sort the list, and the final list is expected to be [1, 2, 3, 4, 5, 6, 7 , 8, .....], then all you need to do is to insert the i+1 to the current position and pop the value of i+1 from its current position. That will reduce the number of iterations you need to sort or swap.
Here's the code that does this with least number of moves. At least that's what I found based on my tests.
def minSwap(ar):
c=0
for i in range(0, len(ar)):
if ar[i] != i+1:
#find the value of i+1 from i+1th position
temp = ar.index(i+1,i+1)
ar.insert(i,i+1) #insert i+1 in the ith position
ar.pop(temp+1) #remove the value of i+1 from the right
c+=1 #every time you do a swap, increment the counter
print (ar) #if you want to check if ar is correct, use this print stmt
return c
a = [1,3,4,5,6,7,2,8]
print (minSwap(a))
The total number of swaps for the above example is 1. It just inserts 2 in the second place and pops out 2 from position 6.
I ran the code for a = [1,6,5,4,3,8,2,7] and it swapped in 5 moves.
I ran the code for a = [1,3,5,4,6,8,2,7] and it swapped in 3 moves.
If you are trying to figure out how this works, use a print statement right after the if statement. It will tell you the element being swapped.
From your code I take it that sorting isn't the issue here, since you know you'll end up with ar[i] == i+1. Given that, why not change your else block to swap the current element into its slot, and repeat until you ar[i] is correct.
else:
while ar[i] != i+1:
temp = ar[i]
ar[i] = ar[temp - 1]
ar[temp - 1] = temp
You don't actually need to do a sort on this array. You just need to figure out the minimum number of swaps needed. If we just look at the following pattern, we can form a hypothesis to be tested:
1234 = 0
1324 = 1, swap 2 and 3
1423 = 2, swap 2 and 4, swap 3 and 4
4213 = 2, swap 1 and 4, swap 3 and 4
4123 = 3, swap 4 and 1, swap 4 and 2, swap 4 and 3
Based on these observations, I think we can work on the hypothesis that the answer will be max(0, n - 1) where n is the count of the number of "out of place" elements.
Then the code becomes simplified to:
def minSwap(ar):
c = 0
for i in range(0, len(ar)):
if ar[i] != i+1:
c = c + 1
return c < 0 ? 0 : c
Note that I don't actually know python so don't know if that last ternary is valid in python.
We have unlimited coins of different values - Calculate the unique combinations of how these coins can make up a specific amount. For example:
n = 4 (say, 4 cents)
coins_list = [1,2] - we have 1-cent coins, and 2-cent coins
The different combinations would be 112, 1111, and 22. (121 and 211 should be excluded since it's not unique - using one 2-cent coin and two 1-cent coin)
I have watched this video: https://www.youtube.com/watch?v=k4y5Pr0YVhg
countless number of times, and edited my codes countless number of times, but I cannot manage to get rid of the same combination of different orders.
def make_change(n, coinlist_index=None):
coin_list = [1, 2]
if coinlist_index == None:
coinlist_index = 0
#coin index position in coin_list; starts at index 0 and cycles through all the coins
if n == 0:
return 1
if n < 0:
return 0
ways = 0
# if I use for i in range(len(coin_list)), it returns an error message saying that index is out of range
for coinlist_index in range(len(coin_list)):
ways += make_change((n - coin_list[coinlist_index]), coinlist_index)
coinlist_index += 1
return ways
make_change(4)
Output: 5
My output was 5 (different ways to make change for 4 cents with 1 and 2-cent coins), instead of 3 (which is what I want).
I'm sure it has to do with the for loop toward the end, but when i change "for coinlist_index in range..." to a different iterator, i, I get an error that says index is out of range.
What is going on, and more importantly, how can I fix it?
EDIT: P.S. This is just a simple example that I'm working through to solve the actual assignment - Which is with 6 types of coins, in cents, (1, 5, 10, 25, 50, 100), and calculate how many ways to make change for 200 dollars. I have seen and tried the dynamic programming method out there, which worked, but we have to use recursion for assignment purposes.
Looks like I got it working. In each recursive pass you want to make sure that you aren't double counting possible ways to make the change. My thought to do this was to make sure that you never go backwards in the coin_list. So for the coin_list [1,2] if we ever use the 2 cent coin we never want the option to use the 1 cent coin afterwards. I made sure it follows this order by changing your for loop a bit:
for i in range(len(coin_list)-coinlist_index):
ways += make_change((n - coin_list[i+coinlist_index-1]), coinlist_index)
In the for loop I subtracted coinlist_index from the upper bound so we don't cycle over all coins once the index reaches 1, then added the index to where you pull from the coin_list, making sure once coinlist_index is 1 or more, we NEVER usecoin_list[0]. This got me to 3 in your sample case, hopefully it works for all cases. Full code:
def make_change(n, coinlist_index=None):
coin_list = [1, 2]
if coinlist_index == None:
coinlist_index = 0
#coin index position in coin_list; starts at index 0 and cycles through all the coins
if n == 0:
return 1
if n < 0:
return 0
ways = 0
# if I use for i in range(len(coin_list)), it returns an error message saying that index is out of range
for i in range(len(coin_list)-coinlist_index):
ways += make_change((n - coin_list[i+coinlist_index-1]), coinlist_index)
coinlist_index += 1
return ways
print(make_change(4))
I feel 5 is actually the correct answer.
1 1 1 1
1 1 2
1 2 1
2 1 1
2 2
Or if you want distinct result, you may store results in the list and remove the duplicate result.
def make_change(n, coinlist_index=0):
coin_list = [1, 2]
if n == 0:
return [[]]
if n < 0:
return []
ways = []
for coinlist_index in range(len(coin_list)):
res = make_change((n - coin_list[coinlist_index]), coinlist_index)
ways += list(map(lambda x : x + [coin_list[coinlist_index]], res))
return ways
def remove_dup(lolist):
res = []
for lst in lolist:
lst.sort()
if lst not in res:
res.append(lst)
return res
print remove_dup(make_change(4))
i try to become familiar with the if-condition statements in haskell
assume that i´ve an argument x and i try the following in haskell
functionname x = if x > 0 then x-5
if x-5 == 0 then 1
else if x-5 /= 0 then functionname x-5
else if x-5 < then 0
so, the idea was to subtract 5 from x, check if the result is 0, if yes, then give a 1.
If not then invoke the function again with the expression x-5.
If the result of x-5 is negative then give a 0.
so, my questions: Would that be correct? Because when i try that, i´ve got a message like parse error on input 'functionname'.
how can i fix that problem? Are the if-else conditions wrong ?
programm :: Int -> Bool
programm x | x > 0 =
if z == 0 then True
else if z < 0 then False
else programm z
where
z = z-2
programm x | x < 0 =
if z == 0 then True
else if z > 0 then False
else programm z
where
z = z+2
so, i wanted to have the possibility to decide of a given number is even. So, i modify your solution a little bit. its the same but, at the beginning of the two declarations i said : x > 0 = .... and x < 0 =...
Because i want to say that for example -4 is also even. for that reason: the first declarations should handle the positive even numbers and the second declarations handles the negative even numbers.
when i give that to the compiler, then the message : Exception appears. Where i ve made the mistake?
Use guards to make things clear:
functionname x
| x > 0 = x - 5
| x - 5 == 0 = 1
| x - 5 /= 0 = functionname (x - 5)
| x - 5 < 0 = 0
Every if needs to have an else clause associated with it.
The very first one doesn't and the very last one doesn't either.
This works just fine:
functionname x = if x > 0 then x-5
else if x-5 == 0 then 1
else if x-5 /= 0 then functionname x-5
else if x-5 < 0 then 0 else 1
so, the idea was to subtract 5 from x, check if the result is 0, if
yes, then give a 1. If not then invoke the function again with the
expression x-5. If the result of x-5 is negative then give a 0.
That might be written like this:
functionname x =
if x' == 0 then 1
else if x' < 0 then 0
else functionname x'
where
x' = x - 5
Here, I use a where clause to locally define x' as x - 5 and then use it for the tests and the recursive call. Your first branch, if x > 0 then x-5, does not appear in your description of what function should do (it gives x - 5 results as result whenever x is larger than zero, which is probably not what you want). Also, note that every if needs an else as well as a then.
so, i wanted to have the possibility to decide of a given number is
even. So, i modify your solution a little bit. its the same but, at
the beginning of the two declarations i said : x > 0 = .... and x < 0
=... Because i want to say that for example -4 is also even. for that reason: the first declarations should handle the positive even numbers
and the second declarations handles the negative even numbers.
First of all, in the second version of your function the definition in the where clause should be z = x + 2, as z = z + 2 will not terminate. This being an evenness test, you also want to perform the tests on x rather than z. With that fixed, your solution with nested conditionals should work fine (note, however, that you are not treating the x == 0 case; the first guard should be x >= 0). There is a more elegant way of writing the function, though:
myEven :: Int -> Bool
myEven x = myEven' (abs x)
where
myEven' x
| x == 0 = True
| x < 0 = False
| otherwise = myEven' (x - 2)
abs is the familiar absolute value function, while myEven' amounts to the x > 0 branch of your original definition. Taking the absolute value of x is the easiest way to avoid writing two nearly equal branches to handle the negative and non-negative cases.
N.B.: While this is probably just a learning exercise, if you ever need to find whether a number is even there is an even function available from the Prelude. There is also mod if you need to test divisibility for other numbers.