I just need a function that removes zeroes from an input list
def no_zero(a):
pos=0
while (pos+1)<=len(a):
if a[pos] == "0":
a.remove[pos]
pos= pos +1
return a
print(no_zero([0,1,0,2,0,3]))
I should be getting an output of 1,2,3 but instead it skips right to return a. Any pointers as to why? Cheers.
You can use a list comprehension:
def no_zero(a):
return [x for x in a if x != 0]
print(no_zero([0,1,0,2,0,3]))
Additionally, the reason your code currently isn't working is because you are comparing the items to a string ("0") instead of an integer (0). You are also attempting to modify a list as you iterate over it, which means that your indices don't correspond to the original indices of the list, and your result will be wrong.
Related
I wish to find Number of sub sequences of length K having total sum S, given an array.
Sample Input:
a=[1,1,1,2,2] & K=2 & S=2
Sample Output:
3 {because a[0],a[1]; a[1]a[2]; a[0]a[2] are only three possible for the case}
I have tried to write a recursive loop in Python for starter but it isn't giving output as expected.Please can you help me find a loophole I might be missing on.
def rec(k, sum1, arr, i=0):
#print('k: '+str(k)+' '+'sum1: '+str(sum1)) #(1) BaseCase:
if(sum1==0 and k!=0): # Both sum(sum1) required and
return 0 # numbers from which sum is required(k)
if(k==0 and sum1 !=0): # should be simultaneously zero
return 0 # Then required subsequences are 1
if(k==0 and sum1==0 ): #
return 1 #
base_check = sum1!=0 or k!=0 #(2) if iterator i reaches final element
if(i==len(arr) and base_check): # in array we should return 0 if both k
return 0 # and sum1 aren't zero
# func rec for getting sum1 from k elements
if(sum1<arr[0]): # takes either first element or rejects it
ans=rec(k-1,sum1,arr[i+1:len(arr)],i+1) # so 2 cases in else loop
print(ans) # i is taken in as iterator to provide array
else: # input to rec func from 2nd element of array
ans=rec(k-1, sum1-arr[0], arr[i+1:len(arr)],i+1)+rec(k, sum1, arr[i+1:len(arr)],i+1)
#print('i: '+str(i)+' ans: '+str(ans))
return(ans)
a=[1,1,1,2,2]
print(rec(2,2,a))
I am still unable to process how to make changes. Once this normal recursive code is written I might go to DP approach accordinlgy.
Using itertools.combinations
Function itertools.combinations returns all the subsequences of a given lengths. Then we filter to keep only subsequences who sum up to the desired value.
import itertools
def countsubsum(a, k, s):
return sum(1 for c in itertools.combinations(a,k) if sum(c)==s)
Fixing your code
Your code looks pretty good, but there are two things that appear wrong about it.
What is this if for?
At first I was a bit confused about if(sum1<arr[0]):. I think you can (and should) always go to the else branch. After thinking about it some more, I understand you are trying to get rid of one of the two recursive calls if arr[0] is too large to be taken, which is smart, but this makes the assumption that all elements in the array are nonnegative. If the array is allowed to contain negative numbers, then you can include a large a[0] in the subsequence, and hope for a negative element to compensate. So if the array can contain negative numbers, you should get rid of this if/else and always execute the two recursive calls from the else branch.
You are slicing wrong
You maintain a variable i to remember where to start in the array; but you also slice the array. Pretty soon your indices become wrong. You should use slices, or use an index i, but not both.
# WRONG
ans=rec(k-1, sum1-arr[0], arr[i+1:len(arr)],i+1)+rec(k, sum1, arr[i+1:len(arr)],i+1)
# CORRECT
ans = rec(k-1, sum1-arr[i], arr, i+1) + rec(k, sum1, arr, i+1)
# CORRECT
ans = rec(k-1, sum1-arr[0], arr[1:]) + rec(k, sum1, arr[1:])
To understand why using both slicing and an index gives wrong results, run the following code:
def iter_array_wrong(a, i=0):
if (a):
print(i, a)
iter_array_wrong(a[i:], i+1)
def iter_array_index(a, i=0):
if i < len(a):
print(i, a)
iter_array_index(a, i+1)
def iter_array_slice(a):
if a:
print(a)
iter_array_slice(a[1:])
print('WRONG')
iter_array_wrong(list(range(10)))
print()
print('INDEX')
iter_array_index(list(range(10)))
print()
print('SLICE')
iter_array_slice(list(range(10)))
Also note that a[i:len(a)] is exactly equivalent to a[i:] and a[0:j] is equivalent to a[:j].
Clean version of the recursion
Recursively count the subsequences who use the first element of the array, and the subsequences who don't use the first element of the array, and add the two counts. To avoid explicitly slicing the array repeatedly, which is an expensive operation, we keep a variable start to remember we are only working on subarray a[start:].
def countsubsum(a, k, s, start=0):
if k == 0:
return (1 if s == 0 else 0)
elif start == len(a):
return 0
else:
using_first_element = countsubsum(a, k-1, s-a[start], start+1)
notusing_first_elem = countsubsum(a, k, s, start+1)
return using_first_element + notusing_first_elem
I want to reverse the string using the Loop & Function. But when I use the following code, it is output the exact same string again. But it suppose to reverse the string. I can't figure out why.
def reversed_word(word):
x=''
for i in range(len(word)):
x+=word[i-len(word)]
print(i-len(word))
return x
a=reversed_word('APPLE')
print(a)
If you look at the output of your debug statement (the print in the function), you'll see you're using the indexes -5 through -1.
Since negative indexes specify the distance from the end of the string, -5 is the A, -4 is the first P, and so on. And, since you're appending these in turn to an originally empty string, you're just adding the letters in the same order they appear in the original.
To add them in the other order, you can simply use len(word) - i - 1 as the index, giving the sequence (len-1) .. 0 (rather than -len .. -1, which equates to 0 .. (len-1)):
def reversed_word(word):
result = ""
for i in range(len(word)):
result += word[len(word) - i - 1]
return result
Another alternative is to realise you don't need to use an index at all since iterating over a string gives it to you one character at a time. However, since it gives you those characters in order, you need to adjust how you build the reversed string, by prefixing each character rather than appending:
def reverse_string(word):
result = ""
for char in word:
result = char + result
return result
This builds up the reversed string (from APPLE) as A, PA, PPA, LPPA and ELPPA.
Of course, you could also go fully Pythonic:
def reverse_string(word):
return "".join([word[i] for i in range(len(word), -1, -1)])
This uses list comprehension to create a list of characters in the original string (in reverse order) then just joins that list into a single string (with an empty separator).
Probably not something I'd hand in for classwork (unless I wanted to annoy the marker) but you should be aware that that's how professional Pythonistas usually tackle the problem.
Let's say your word is python.
You loop will then iterate over the values 0 through 5, since len(word) == 6.
When i is 0, i-len(word) is -6 (note carefully that this value is negative). You'll note that word[-6] is the character six places to the left from the end of the string, which is p.
Similarly, when i is 1, i-len(word) is -5, and word[i-len(word)] is y.
This pattern continues for each iteration of your loop.
It looks like you intend to use positive indices to step backward through the string with each iteration. To obtain this behavior, try using the expression len(word)-i-1 to index your string.
def reversed_word(word):
reversed = ''
for i in range(len(word)-1, -1, -1):
reversed += word[i]
return reversed
print(reversed_word("apple"))
I am supposed to ask the user to input a number then print all the numbers preceding that but, I have to insert a 0 after every even number.
I am a beginner in Python so, I don't know much.
n=0
length_even=[]
length_odd=[]
def length_list(n):
length=int(input('Enter the length of the list: '))
for i in range(1,length+1):
if(i%2 != 0):
length_odd.append(i)
elif(i%2 == 0) :
length_even.append(i)
length_even.insert(i,0)
total=length_odd+length_even
return total
length_list(n)
If a user enters 5, I expect the output to be [1,2,0,3,4,0,5] but, the actual output is (1,3,5,2,0,4,0].
The problem is that you create two independent lists (length_even and length_odd) and concatenate (+ operator) them after you have added the values to each list (length_odd contains: 1,3,5 and length_even contains: 2,0,4,0). The result you get is therefore: 1,3,5,2,0,4,0. Why don't you just use one list? Have a look at the code below (I made only slight changes to avoid confusion):
n=0
lengthList = []
def length_list(n):
length=int(input('Enter the length of the list: '))
for i in range(1,length+1):
if(i%2 != 0):
lengthList.append(i)
elif(i%2 == 0) :
lengthList.append(i)
lengthList.append(0)
return lengthList
length_list(n)
Im trying to see if a value in one array is equal to that in another array, the values are integer values.
Ive tried turning them into string and integers from the array but get the error that they cannot be converted implicitly.
winningnumber = []
usernumber = []
print(winningnumber)
print(usernumber)
if(winningnumber == usernumber):
print("Exact number")
I would then get an output like so
[1]
['1']
In order to do this, what you want to do is access the first item of each array, and compare that value.
There are a lot of ways to do this, but here is a little driver program to show you one way.
# Defining a function to see if they match
def is_winning(arr1, arr2):
# Grabbing the first element in each array
# denoted by the [0], for the "0th" element
arr1_first_ele = arr1[0]
arr2_first_ele = arr2[0]
# If the first element in the first array matches the first element in the second
if arr1_first_ele == arr2_first_ele:
# Print out they match
print("They match")
# Otherwise
else:
# Print out that they dont
print("They don't match")
def main():
# Example arrays
test_array_one = [1,3,4]
test_array_two = [5,4,3]
# This should print out "They don't match"
is_winning(test_array_one, test_array_two)
# Example arrays
test_array_three = [6,7,8]
test_array_four = [6,5,4]
# This should print out "They match"
is_winning(test_array_three, test_array_four)
main()
This evaluates to:
They don't match
They match
I've built a function that checks for repeats of a specific number in a list named xs. V is the number to check for repeats of. It needs to return True if there are more than one occurrences of the number and if there are none, it needs to return False.
I'm failing one test case which is input xs=[1,2,1] v=1, this function needs to return True, but my code is making it False. Can you see where I went wrong?
Here is my current code:
def has_repeat(xs, v):
count=0
for num in range(len(xs)):
if num == v:
count+=1
if count>1:
return True
else:
return False
You're actually iterating over the range of the length of the list, not the items in the list.
The range function returns a list of numbers from 0 (by default) to the number you provide, in this case 3 (not inclusive). See Python documentation.
As an example if you try:
l = [1, 2, 3]
print(range(len(l)))
It will print out [0, 1, 2]
What you should do is instead of
for num in range(len(xs))
do
for num in xs:
You can try it out on PyFiddle here
As an added tasty bonus, you could change this to use the .count method on your list of items to check how many occurrences of that number are in the list, removing the need to iterate the list at all, like so:
count = xs.count(v)