[python] Check element exists in array

In PHP there a function called isset() to check if something (like an array index) exists and has a value. How about Python?

I need to use this on arrays because I get "IndexError: list index out of range" sometimes.

I guess I could use try/catching, but that's a last resort.

has_key is fast and efficient.

Instead of array use an hash:

valueTo1={"a","b","c"}

if valueTo1.has_key("a"):
        print "Found key in dictionary"

EAFP vs. LBYL

I understand your dilemma, but Python is not PHP and coding style known as Easier to Ask for Forgiveness than for Permission (or EAFP in short) is a common coding style in Python.

See the source (from documentation):

EAFP - Easier to ask for forgiveness than permission. This common Python coding style assumes the existence of valid keys or attributes and catches exceptions if the assumption proves false. This clean and fast style is characterized by the presence of many try and except statements. The technique contrasts with the LBYL style common to many other languages such as C.

So, basically, using try-catch statements here is not a last resort; it is a common practice.

"Arrays" in Python

PHP has associative and non-associative arrays, Python has lists, tuples and dictionaries. Lists are similar to non-associative PHP arrays, dictionaries are similar to associative PHP arrays.

If you want to check whether "key" exists in "array", you must first tell what type in Python it is, because they throw different errors when the "key" is not present:

>>> l = [1,2,3]
>>> l[4]

Traceback (most recent call last):
  File "<pyshell#2>", line 1, in <module>
    l[4]
IndexError: list index out of range
>>> d = {0: '1', 1: '2', 2: '3'}
>>> d[4]

Traceback (most recent call last):
  File "<pyshell#6>", line 1, in <module>
    d[4]
KeyError: 4

And if you use EAFP coding style, you should just catch these errors appropriately.

LBYL coding style - checking indexes' existence

If you insist on using LBYL approach, these are solutions for you:

• for lists just check the length and if possible_index < len(your_list), then your_list[possible_index] exists, otherwise it doesn't:

  >>> your_list = [0, 1, 2, 3]
  >>> 1 < len(your_list) # index exist
  True
  >>> 4 < len(your_list) # index does not exist
  False
  

• for dictionaries you can use in keyword and if possible_index in your_dict, then your_dict[possible_index] exists, otherwise it doesn't:

  >>> your_dict = {0: 0, 1: 1, 2: 2, 3: 3}
  >>> 1 in your_dict # index exists
  True
  >>> 4 in your_dict # index does not exist
  False
  

Did it help?

You may be able to use the built-in function dir() to produce similar behavior to PHP's isset(), something like:

if 'foo' in dir():  # returns False, foo is not defined yet.
    pass

foo = 'b'

if 'foo' in dir():  # returns True, foo is now defined and in scope.
   pass

dir() returns a list of the names in the current scope, more information can be found here: http://docs.python.org/library/functions.html#dir.

In Python you might be in for some surprises if you ask for forgiveness in this case.

try-except is not the right paradigm here.

If you accidentally get negative indices your in for a surprise.

Better solution is to provide the test function yourself:

def index_in_array(M, index):
    return index[0] >= 0 and index[1] >= 0 and index[0]< M.shape[0] and index[1] < M.shape[1]

`e` in ['a', 'b', 'c']  # evaluates as False
`b` in ['a', 'b', 'c']  # evaluates as True

EDIT: With the clarification, new answer:

Note that PHP arrays are vastly different from Python's, combining arrays and dicts into one confused structure. Python arrays always have indices from 0 to len(arr) - 1, so you can check whether your index is in that range. try/catch is a good way to do it pythonically, though.

If you're asking about the hash functionality of PHP "arrays" (Python's dict), then my previous answer still kind of stands:

`baz` in {'foo': 17, 'bar': 19}  # evaluates as False
`foo` in {'foo': 17, 'bar': 19}  # evaluates as True