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Python noob question. Looping through all to find correct combination

Sorry for noob question. Just started learning coding with Python - beginners level.
Wasn't able to find in the net what I require.
Creating function to loop through the entire line, in order to find right combination - 7,8,9 - regardless of input length and of target position, and return 'true' if found. Wasn't able to devise the function correctly. Not sure how to devise function clearly and at all this far.
Your help is much appreciated.
This is what I came up with so far (not working of course):
def _11(n):
for loop in range(len(n)):
if n[loop]==[7,8,9]:
return True
else:
return False
print(_11([1000,10,11,34,67,89,334,5567,6534,765,2,3,5,6,112,7,8,9,11111]))
It always returns False. Tried with (*n) to no avail.

The answer offered by #Carson is entirely correct.
I offer this not really as an answer to the question but as an alternative and more efficient approach.
In OP's question he is looking for an occurrence of 3 consecutive values described by way of a list. Let's call that a triplet.
If we iterate over the input list one element at a time we create lots of triplets before comparing them.
However, we can make this more efficient by searching the input list for any occurrence of the first item in the target triplet. In that way we are likely to slice the input list far less often.
Here are two implementations with timings...
from timeit import timeit
def _11(n, t):
offset = 0
lt = len(t)
m = len(n) - lt
while offset < m:
try:
offset += n[offset:].index(t[0])
if n[offset:offset+lt] == t:
return True
offset += 1
except ValueError:
break
return False
def _11a(n, t):
for index in range(len(n) - len(t)):
if n[index:index + len(t)] == t:
return True
return False
n = [1000,10,11,34,67,89,334,5567,6534,765,2,3,5,6,112,7,8,9,11111]
t = [7, 8, 9]
for func in _11, _11a:
print(func.__name__, timeit(lambda: func(n, t)))
Output:
_11 0.43439731000012216
_11a 1.8685798310000337

There are two mistakes with your code.
Indexing into a loop returns 1 element, not multiple. When you write n[loop], you're getting 1 value, not a list.
You shouldn't return false that early. Your code exits after the first step in the loop, but it should go through the entire loop before returning false.
Consider the following snippet:
def has_subarr(arr, subarr):
"""Tests if `subarr` exists in arr"""
for i in range(len(arr)):
if arr[i:i+len(subarr)] == subarr:
return True
return False
This code is more general than your example, it accepts the value to check for as another argument. Notice the use of : in the array access. This allows you to return multiple elements in an array. Also notice how the return False is only reached once the entire loop has completed.

First, n[loop] return a single element, not a sublist. You should use n[loop+3]. But this will introduce a problem where loop+3 exceeds the length of the list. So the solution may be:
def _11(n):
for loop in range(len(n)-3):
if n[loop:loop+3]==[7,8,9]:
return True
else:
return False
print(_11([1000,10,11,34,67,89,334,5567,6534,765,2,3,5,6,112,7,8,9,11111]))
Your actual code return during the first iteration. You only test once. You must modify the indentation as in:
def _11(n):
target = [7,8,9]
for index in range( len(n) - len(target)):
if n[index:index + len(target)] == [7,8,9]:
return True
return False
print(_11([1000,10,11,34,67,89,334,5567,6534,765,2,3,5,6,112,7,8,9,11111]))

You can try checking the str representation of the 2 lists:
import re
def _11(n):
if re.search("(?<![0-9-.'])7, 8, 9(?![0-9.])",str(n)):
return True
return False
print(_11([27,8,9]))
The Output:
False

Check if an array is sorted or not using recursion

Here is my code:
def isSorted(arr):
n = len(arr)
if n == 0 or n == 1:
return True
elif arr[n-2] > arr[n-1]:
return False
return isSorted(arr[0:n-1])
arr = [1,9,9,4,5]
isSorted(arr)
if isSorted:
print("yes")
else:
print("no")
Answer is always yes, even if the array is unsorted. Can anybody please explain what mistake am I making?

Your code was fine with the exception of the error noted in the comments: you were ignoring the return value of isSorted(arr) by just checking if isSorted:. isSorted is a callable (and not None or zero or an empty string, etc.) and thus evaluates to True.
Here's a slight modification to your code, using negative indices to count back from the end of the array. For example, -2 is the same as n-2 when n is the length of the array.
I've also thrown in a little syntactic sugar at the end, using python's ternary operator.
def isSorted(arr):
if len(arr) < 2:
return True
elif arr[-2] > arr[-1]:
return False
return isSorted(arr[:-1])
arr = [1,9,9,4,5]
print("yes" if isSorted(arr) else "no")

A recursive approach can be implemented like this:
def isSorted(lst):
if len(lst) < 2:
return True
return False if lst[1] < lst[0] else isSorted(lst[1:])
However, it's going to be slow compared with:
lst == sorted(lst)
...due to the efficient implementation of the sort() function

As others have pointed out, there are non-recursive solutions which will outperform recursive implementations.
Having said that, if you insist on using recursion you should seek approaches that will not give you a stack overflow with larger arrays. Solutions which whittle the size of the problem down one element at a time will cause stack overflow on arrays that are larger than ~1000 elements. To avoid this, use an approach which cuts the problem size in half at each level of the recursion. The following algorithm does this:
def isSorted(ary):
if len(ary) < 2:
return True
mid = len(ary) // 2
return (ary[mid-1] <= ary[mid]) and isSorted(ary[:mid]) and isSorted(ary[mid:])
The logic is:
Arrays of length less than 2 are trivially considered sorted, so return True if this is the case;
Otherwise, find the mid-range value which will split the array into two sub-arrays which differ in length by no more than one;
The array is sorted if and only if the last element of the first sub-array is ≤ the first element of the second subarray, and both sub-arrays are sorted.
Since and is used, short-circuiting can occur. The recursive stack size is O(log(len(ary)).
As Mark pointed out, the reason you always get "yes" is because isSorted, without an argument, is a reference to the function and is considered "truthy" in python.

def isSorted(arr):
n = len(arr)
if n == 0 or n == 1:
return True
elif arr[n-2] > arr[n-1]:
return False
return isSorted(arr[0:n-1])
arr = [1,9,9,4,5]
#isSorted(arr)
#You should use "if isSorted(arr):" instead of "isSorted(arr)".
if isSorted(arr):
print("yes")
else:
print("no")

Simply add the parameters to the function in the if statement:
if isSorted(arr):
print("yes")
else
print("no")

Leetcode jumpgame recursive approach

Given the following problem:
You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.
Return true if you can reach the last index, or false otherwise.
Example 1:
Input: nums = [2,3,1,1,4]
Output: True
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [3,2,1,0,4]
Output: False
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
I am trying to come up with a recursive solution. This is what I have so far. I am not looking for the optimal solution. I am just trying to solve using recursion for now. If n[i] is 0 I want the loop to go back to the previous loop and continue recursing, but I can't figure out how to do it.
def jumpGame(self, n: []) -> bool:
if len(n) < 2:
return True
for i in range(len(n)):
for j in range(1, n[i]+1):
next = i + j
return self.jumpGame(n[next:])
return False

If you want to do recursively and you said no need to be optimal ( so not memoized ), you could go with the below method. You don't need nested loops.
Also no need to explore all paths, you could optimize by looking at the step that you are going by checking i + (jump) < n
def jumpGame(a, i):
if i > len(a) - 1:
return False
if i == len(a) - 1:
return True
reached = False
for j in range(1, a[i] + 1):
if i + j < len(a):
reached = jumpGame(a, i + j)
if reached:
return True
return reached
print(jumpGame([2, 3, 1, 1, 4], 0))
print(jumpGame([3,2,1,0,4], 0))
True
False

When considering recursive solutions, the first thing you should consider is the 'base case', followed by the 'recursive case'. The base case is just 'what is the smallest form of this problem for which I can determine an answer', and the recursive is 'can I get from some form n of this problem to some form n - 1'.
That's a bit pedantic, but lets apply it to your situation. What is the base case? That case is if you have a list of length 1. If you have a list of length 0, there is no last index and you can return false. That would simply be:
if len(ls) == 0:
return False
if len(ls) == 1:
return True
Since we don't care what is in the last index, only at arriving at the last index, we know these if statements handle our base case.
Now for the recursive step. Assuming you have a list of length n, we must consider how to reduce the size of the problem. This is by making a 'jump', and we know that we can make a jump equal to a length up to the value of the current index. Then we just need to test each of these lengths. If any of them return True, we succeed.
any(jump_game(n[jump:] for jump in range(1, n[0] + 1)))
There are two mechanisms we are using here to make this easy. any takes in a sequence and quits as soon as one value is True, returning True. If none of them are true, it will return False. The second is a list slice, n[jump:] which takes a slice of a list from the index jump to the end. This might result in an empty list.
Putting this together we get:
def jump_game(n: list) -> bool:
# Base cases
if len(n) == 0:
return False
if len(n) == 1:
return True
# Recursive case
return any(jump_game(n[jump:]) for jump in range(1, n[0] + 1))
The results:
>>> jump_game([2,3,1,1,4])
True
>>> jump_game([3,2,1,0,1])
False
>>> jump_game([])
False
>>> jump_game([1])
True
I'm trying to lay out the rigorous approach here, because I think it helps to clarify where recursion goes wrong. In your recursive case you do need to iterate through your options - but that is only one loop, not the two you have. In your solution, in each recursion, you're iterating (for i in range(len(n))) through the entire list. So, you're really hiding an iterative solution inside a recursive one. Further, your base case is wrong, because a list of length 0 is considered a valid solution - but in fact, only a list of length 1 should return a True result.
What you should focus on for recursion is, again, solving the smallest possible form(s) of the problem. Here, it is if the list is one or zero length long. Then, you need to step each other possible size of the problem (length of the list) to a base case. We know we can do that by examining the first element, and choosing to jump anywhere up to that value. This gives us our options. We try each in turn until a solution is found - or until the space is exhausted and we can confidently say there is no solution.

I want to check if a string is abecedarian

I want to check if a string is abecedarian (in alphabetical order).
def is_abecdarian(s):
index = 0
while index < len(s)-1:
if s[index + 1] > s[index]:
return True
index += 1
return False
print(is_abecdarian('babcd'))
If I switch my False and True around and switch the greater than to less than it works. What I do not understand is 'a' is not greater than 'b', yet when I run the above code it returns True.

The return True statement exits the function, so right now it runs until there are two letters that are in a correct order and then exits.

The way your loop is written, it will return True if any pair of characters is ordered. It only returns False if all pairs fail. The first pair ba fails the if check, but the second pair ab passes and the function returns True.
This isn't right, because if you think about it, you can't know that a string is abecedarian until you've examined all of the characters. You should only be returning True when you're done with the loop. Conversely, as soon as you find a pair of characters that's out of order, it is safe to return False.
That's why flipping the booleans and the < operation works.

You need to return False only if next character is < than current.
If there were no such an occurrence during all the loop then return True.
Here is correct function:
def is_abecdarian(s):
index = 0
while index < len(s)-1:
if s[index + 1] < s[index]:
return False
index += 1
return True
print(is_abecdarian('babcd'))
print(is_abecdarian('abcz'))

Why it's not working has already been explained, so, only to add to what has been explained:
def is_abecdarian(s):
return all(s[i] < s[i+1] for i in range(len(s)-1))
print(is_abecdarian('babcd'))
print(is_abecdarian('ejmpq'))
Where
all(argument) returns True only if all boolean in the argument are True;
s[i] < s[i+1] is a comparison between the i-th element of s to the next one.
s[i] < s[i+1] for i in range(len(s)-1) makes the comparison for all pairs of consecutive elements of s.

How to write more Pythonic Code

I started learning python today from the tutorial on the official site.
When reading about filter(function, sequence) i thought of making a function that returns if a number is prime to use it with the filter.
notDividedBy = [2,3,4,5,6,7,8,9]
def prime(num):
"""True if num is prime, false otherwise"""
copy = notDividedBy[:]
check = True
if num in copy:
copy.remove(num)
for x in copy:
if num % x == 0:
check = False
break
return check
The above code works in the shell.
My question is: Since i feel like although a solution, it is not the most elegant one, can anyone transform this code to something more python-like?(better structure? less lines?)
I believe it would help me for better understanding of the basics of the language.
The thing is, don't use any imports or anything, just simple staff.

Creating many many copies of lists is not a particularly efficient way of doing things. Instead use the xrange() (Python 2.x) or range() (Python 3) iterator. Here's one (naive) way you could implement a primality test:
from math import sqrt
def isPrime(n):
if n < 2: return False
if n == 2: return True
if not n % 2: return False #test if n is even
#we've already remove all the even numbers, no need to test for 2
#we only need to test up to sqrt(n), because any composite numbers can be
# factored into 2 values, at least one of which is < sqrt(n)
for i in xrange(3, int(sqrt(n)) + 1, 2):
if not n % i:
return False
return True

How about this one:
def is_prime(num):
return not any(num%i == 0 for i in xrange(2,num/2+1))
for i in xrange(10):
print i, is_prime(i)
Explanation
start with:
(num%i==0 for i in xrange(2,num/2+1))
This is a generator expression. I could have made it a list comprehension:
[num%i==0 for i in xrange(2,num/2+1)]
The list comprehension is equivalent to:
ll=[]
for i in xrange(2,num/2+1):
ll.append(num%i==0)
The difference between the generator and the list comprehension is that the generator only gives up it's elements as you iterate over it -- whereas the list comprehension calculates all the values up front. Anyway, from the above code, you can see that the expression generates a sequence of True's and False's. True if the number can be divided by i and False otherwise. If we generate a sequence of all False numbers, we know we have a prime.
The next trick is the any built in function. It basically searches through an iterable and checks if any of the values is True. As soon as it hits a True, it returns True. If it gets to the end of the iterable, it returns False. So, if the entire sequence is False (a prime number) then any will return False, otherwise it returns True. This would be perfect for a not_prime function, but our function is is_prime, so we just need to invert that result using the not operator.
The benefit of using the generator expression is that it is nice and concise, but also that it allows any to return before checking every value which means that as soon as it finds a number that divides num, it returns instead of generating all num/2 numbers.
Anyway, I hope this explanation is helpful. If not, feel free to leave a comment and I'll try to explain better.

One thing off the bat, if you are going to implement prime testing in this fashion, there's no reason to use an auxillary array
def prime(num):
"""True if num is prime, false otherwise"""
check = True
#if num in copy:
# copy.remove(num)
for x in range(2,x-1):
if num % x == 0:
check = False
break
return check

Here's a 2 liner using filter().
def prime(num):
"""True if num is prime, false otherwise"""
if num < 2:
return False
return len(filter(lambda x: num % x == 0, range(2, num))) == 0