## Scheme: How to check if all elements of a list are identical

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I'd like to create a Scheme function that yields true if it is passed a list that is composed entirely of identical elements. Such a list would be '(1 1 1 1). It would yield false with something like '(1 2 1 1).

This is what I have so far:

```    (define (list-equal? lst)
(define tmp (car lst))
(for-each (lambda (x)
(equal? x tmp))
lst)
)
```

Clearly this is incorrect, and I'm new to this. I guess I'm unable to express the step where I'm supposed to return `#t` or `#f`.

EDIT: I fiddled a bit and found a solution that seems to work very well, and with a minimal amount of code:

```(define (list-equal? lst)
(andmap (lambda (x)
(equal? x (car lst)))
lst))
```

Thanks again for the help everyone.

Minimal amount of code, if you don't care that it only works for numbers:

```(define (list-equel? lst)
(apply = lst))
```

Examples:

```> (list-equel? '(1 1 2 1))
#f
> (list-equel? '(1 1 1 1))
#t
> (list-equel? '(1))
#t
```

Python : Check if all elements in a List are same or matches a , How do you check if all elements in a list are the same? Given a list, write a Python program to check if all the elements in given list are same. Example: Input: ['Geeks', 'Geeks', 'Geeks', 'Geeks', ] Output: Yes Input: ['Geeks', 'Is', 'all', 'Same', ] Output: No. There are various ways we can do this task. Let’s see different ways we can check if all elements in a List are same.

The `andmap` solution is nice, but if `andmap` is not available, you can use this. It uses basic operations (and, or, null check, equality check) and handles empty lists and one element lists. Similar to Sean's implementation, but no helper definition is necessary.

```(define (list-equal? args)
(or (or (null? args)
(null? (cdr args)))
(and (eq? (car args) (cadr args))
(list-equal? (cdr args)))))
```

Determine If All Elements Are the Same in a Java List, How do you check if all elements in a list are equal Java? Method 3 : Using sum () + zip () + len () Using sum () + zip (), we can get sum of one of the list as summation of 1 if both the index in two lists have equal elements, and then compare that number with size of other list. This also requires first to check if two lists are equal before this computation.

Try something like this:

```(define (list-equal? lst)
(define (helper el lst)
(or (null? lst)
(and (eq? el (car lst))
(helper (car lst) (cdr lst)))))
(or (null? lst)
(helper (car lst) (cdr lst))))
```

This might not be the cleanest implementation, but I think it will correctly handle the cases of empty lists and one-element lists.

How can I compare two ordered lists in python?, How do you check if two lists are equal in Python? Some list operations require us to check if all the elements in the list are unique. This usually happens when we try to perform the set operations in a list. Hence this particular utility is essential at these times. Lets discuss certain methods by which this can be performed. Method #1 : Naive Method.

In R6RS there's the `for-all` function, which takes a predicate and a list, and returns `#t` if the predicate returns true for all elements in the list and `#f` otherwise, which is exactly what you need here.

So if you're using R6RS (or any other scheme dialect that has the `for-all` function), you can just replace `for-each` with `for-all` in your code and it will work.

My other car is a cdr - Aleksandra Sikora, or not, a == b should give you true / false with ordering taken into account. Here, c will contain an array with 3 elements all of which are true (for your example). One simple solution is to use two nested loops. For every element, check if it repeats or not. If any element repeats, return false. If no element repeats, return false. An efficient solution is to Hashing. We put all array elements in a HashSet. If size of HashSet remains same as array size, then we return true. C++.

```(define (list-equal? lst)
(if (= (cdr lst) null)
true
(and (equal? (car lst) (cadr lst))
(list-equal? (cdr lst)))))
```

Python, is an acronym from the phrase Contents of the Decrement part of the Register. Python : Check if a list contains all the elements of another list; Python: check if two lists are equal or not ( covers both Ordered & Unordered lists) Check if all elements in a list are None in Python; Python : Get number of elements in a list, lists of lists or nested list; Python: Check if all values are same in a Numpy Array (both 1D and 2D)

Python, Another method is to take the first element and multiply it with the length of given list to form a new list, So that the new list contains identical elements to first elements of given list size, and then compare it with the given list. Write a Scheme function DEL-LELEMENT that takes a list as a parameter and returns a list identical to the parameter except the last element has been deleted. For example, (DEL-LELEMENT '(8 2 3 7 6)) should return (8 2 3 7) *Please explain your code and submit the source code.

Check if list contains a value, in Scheme, Given a list, write a Python program to check if all the elements in given list are same. Equal. But In Python, we can do the same task in much interesting ways. Pairs are not mutable (but see Mutable Pairs and Lists). A list is recursively defined: it is either the constant null, or it is a pair whose second value is a list. A list can be used as a single-valued sequence (see Sequences). The elements of the list serve as elements of the sequence. See also in-list.

MIT Scheme Reference, Idiom #12 Check if list contains a value list is an iterable finite container. Scheme. (define (contains list x) (if (null? list) #f) (if (equal? (car list) x) #t) (else

• Though you don't even need the length condition. `(equal?)` with no arguments => `#t`.
• Fun. When I tested the idea, I subconsciously corrected it to `equal?`. I think I'll fix that for @aaz.
• @JasonFruit: In standard Scheme, `equal?` is not variadic the way `=` is. :-) (I tested with Racket, for example: "equal?: expects 2 arguments, given 3: 1 2 3") You're going to have to use folding, sorry. :-P
• `(or 1 2 3)` is also legal.
• This would work, with the added step that you still must check that `lst` is not empty.
• @Sean: Yes, good point. That or get rid of the `tmp` variable and just call `(car lst)` every time in the predicate (which would never get called for the empty list, avoiding the problem).