Python check if list items are integers?

Fast, simple, but maybe not always right:

>>> [x for x in mylist if x.isdigit()]
['1', '2', '3', '4']

More traditional if you need to get numbers:

new_list = []
for value in mylist:
    try:
        new_list.append(int(value))
    except ValueError:
        continue

Note: The result has integers. Convert them back to strings if needed, replacing the lines above with:

try:
    new_list.append(str(int(value)))

You can use exceptional handling as str.digit will only work for integers and can fail for something like this too:

>>> str.isdigit(' 1')
False

Using a generator function:

def solve(lis):                                        
    for x in lis:
        try:
            yield float(x)
        except ValueError:    
            pass

>>> mylist = ['1','orange','2','3','4','apple', '1.5', '2.6']
>>> list(solve(mylist))                                    
[1.0, 2.0, 3.0, 4.0, 1.5, 2.6]   #returns converted values

or may be you wanted this:

def solve(lis):
    for x in lis:
        try:
            float(x)
            return True
        except:
            return False
...         
>>> mylist = ['1','orange','2','3','4','apple', '1.5', '2.6']
>>> [x for x in mylist if solve(x)]
['1', '2', '3', '4', '1.5', '2.6']

or using ast.literal_eval, this will work for all types of numbers:

>>> from ast import literal_eval
>>> def solve(lis):
    for x in lis:
        try:
            literal_eval(x)
            return True
        except ValueError:   
             return False
...         
>>> mylist=['1','orange','2','3','4','apple', '1.5', '2.6', '1+0j']
>>> [x for x in mylist if solve(x)]                               
['1', '2', '3', '4', '1.5', '2.6', '1+0j']

Try this:

mynewlist = [s for s in mylist if s.isdigit()]

---

From the docs:

str.isdigit()

Return true if all characters in the string are digits and there is at least one character, false otherwise.

For 8-bit strings, this method is locale-dependent.

---

As noted in the comments, isdigit() returning True does not necessarily indicate that the string can be parsed as an int via the int() function, and it returning False does not necessarily indicate that it cannot be. Nevertheless, the approach above should work in your case.

The usual way to check whether something can be converted to an int is to try it and see, following the EAFP principle:

try:
    int_value = int(string_value)
except ValueError:
    # it wasn't an int, do something appropriate
else:
    # it was an int, do something appropriate

So, in your case:

for item in mylist:
    try:
        int_value = int(item)
    except ValueError:
        pass
    else:
        mynewlist.append(item) # or append(int_value) if you want numbers

---

In most cases, a loop around some trivial code that ends with mynewlist.append(item) can be turned into a list comprehension, generator expression, or call to map or filter. But here, you can't, because there's no way to put a try/except into an expression.

But if you wrap it up in a function, you can:

def raises(func, *args, **kw):
    try:
        func(*args, **kw)
    except:
        return True
    else:
        return False

mynewlist = [item for item in mylist if not raises(int, item)]

… or, if you prefer:

mynewlist = filter(partial(raises, int), item)

---

It's cleaner to use it this way:

def raises(exception_types, func, *args, **kw):
    try:
        func(*args, **kw)
    except exception_types:
        return True
    else:
        return False

This way, you can pass it the exception (or tuple of exceptions) you're expecting, and those will return True, but if any unexpected exceptions are raised, they'll propagate out. So:

mynewlist = [item for item in mylist if not raises(ValueError, int, item)]

… will do what you want, but:

mynewlist = [item for item in mylist if not raises(ValueError, item, int)]

… will raise a TypeError, as it should.