为什么会出现语法错误（python数字检查功能）？

Question

I've written this function in Python, which is designed to check if any list element is a number, and return that list element if it is. This is the code:

def check_for_number(list):
x = 0
print(isinstance(list[x], (int, float))
true_or_false = False
for x in range(len(list)-1):
    if isinstance(list[x], (int, float) == True):
        true_or_false = True
        num = list[x]
    x += 1
print(true_or_false)
return true_or_false 
return num

Whenever I try to run it I get a syntax error saying that the colon at the end of the if statement is an 'unexpected token', and every item in the last two lines of the if statement gives the same unexpected token error. I've checked the indentation and I can't see anything wrong with it, what am I missing? Thanks.

Answer 1

You just have to indend the code of your function and fix the if isinstance(list[x], (int, float) == True): part and close the paranthesis of your first print statement.

def check_for_number(list):
    x = 0
    print(isinstance(list[x], (int, float)))
    true_or_false = False
    for x in range(len(list)-1):
        if isinstance(list[x], (int, float)) == True:
            true_or_false = True
            num = list[x]
        x += 1
    print(true_or_false)

    # Decide what you want to return here
    return true_or_false 
    return num

If you are interested in improving your code, remove the == True part, as has been stated in the comments as well. And from your question I assume you want to return fals, if true_or_false is False or num otherwise.

If you add a break statement in the loop, the loop will be exited when you have found the first number. So your computer does not need to loop through the complete list and this can save you some execution time.

I also expect, your x += 1 statement is not what you want to do. The for ... in range ... loop will increase your x in each cycle. That is why x += 1 will make your code skip every second list element. You also will not need to declare x first.

def check_for_number(list):
    print(isinstance(list[x], (int, float)))
    true_or_false = False
    for x in range(len(list)-1):
        if isinstance(list[x], (int, float)):
            true_or_false = True
            num = list[x]
            break
    print(true_or_false)

    if (true_or_false):
        return num
    else:
        return False

---

Concering your question about the unnecessary == True part:

The if statement is working like the following pseudo code.

if (CONDITION) == True then do something special

So, if you add a == True , python would check it like so:

if (valeu == True) == True then do something special

which is the same as:

if (value) == True then do something special

Answer 2

Here is a solution using a list comprehension. It checks each element against the Number abstract base class. It will return a list of numbers, since there might be more than 1 numeric elelement.

import numbers

def get_numbers(l):
    return [x for x in l if isinstance(x, numbers.Number)]

example:

>>> l = ['1', 4.0, 'abc', 'XYZ', 'test', 5]
>>> nums = get_numbers(l)
>>> print(nums)
[4.0, 5]

Answer 3

Python is tab-sensitive and intolerant of errors in this regard. I'm also seeing an instance of an unmatched parenthesis on line 6. Try the following and you might get a more informative error on what to fix next:

def check_for_number(list): 
    x = 0 
    print(isinstance(list[x], (int, float))
    true_or_false = False 
    for x in range(len(list)-1):
        if isinstance(list[x], (int, float) == True): # missing a parenthesis to close isinstance. 
        # "== True" is unnecessary because isinstance() will resolve to True or False
            true_or_false = True 
            num = list[x]
            x += 1 
            print(true_or_false) 
            return true_or_false 
    return num # this will result in unexpected behavior...
        # ...because num will not be defined if this line is reached.
        # Either the return inside the if will finish the method or num will never be defined.

It's a bit ambiguous how some of that should be indented because I can't tell what you're trying to do.

If you're trying to return to values, consider returning a list or a dictionary containing the values.