What is the python equivalent of JavaScript's Array.prototype.some / every?

Question

Does python have any equivalent to JavaScript's Array.prototype.some / every?

Trivial JavaScript example:

var arr = [ "a", "b", "c" ]; arr.some(function (element, index) {     console.log("index: " + index + ", element: " + element)     if(element === "b"){         return true;     } }); 

Will output:

index: 0, element: a index: 1, element: b 

The below python seems to be functionally equivalent, but I do not know if there is a more "pythonic" approach.

arr = [ "a", "b", "c" ] for index, element in enumerate(arr):     print("index: %i, element: %s" % (index, element))     if element == "b":         break 

Answer 1

Python has all(iterable) and any(iterable). So if you make a generator or an iterator that does what you want, you can test it with those functions. For example:

some_is_b = any(x == 'b' for x in ary) all_are_b = all(x == 'b' for x in ary) 

They are actually defined in the documentation by their code equivalents. Does this look familiar?

def any(iterable):     for element in iterable:         if element:             return True     return False 

Answer 2

No. NumPy arrays have, but standard python lists don't. Even so, the numpy array implementations are not what you'd expect: they don't take a predicate, but evaluate every element by converting them to boolean.

Edit: any and all exist as functions (not as methods), but they don't apply predicates, but consider booleanized values as numpy methods.

In Python, some could be:

def some(list_, pred):     return bool([i for i in list_ if pred(i)])  #or a more efficient approach, which doesn't build a new list def some(list_, pred):     return any(pred(i) for i in list_) #booleanize the values, and pass them to any 

You could implement every:

def every(list_, pred):     return all(pred(i) for i in list_) 

Edit: dumb sample:

every(['a', 'b', 'c'], lambda e: e == 'b') some(['a', 'b', 'c'], lambda e: e == 'b') 

Try them by urself