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