Shortening if statements Python

Is there a way to shorten this:

if str(num[p]).find('0') == 
int(-1) and
str(num[p]).find('1') == int(-1) 
and str(num[p]).find('2') == 
int(-1) and str(num[p]).find('3') 
== int(-1) and 
str(num[p]).find('4') == int(-1) 
and str(num[p]).find('5') == 
int(-1) and str(num
[p]).find('6') == int(-1) and 
str(num[p]).find('7') == int(-1) 
and str(num[p]).find('8') == 
int(-1) and str(num[p]).find('9') 
== int(-1) and str
(num[p]).find('.') == int(-1):

Note:the num[p] is part of a for loop and is not an error.

>Solution :

I’d write this as:

if not any(c in num[p] for c in "0123456789."):

Note that if num[p] is already a string, the str() is not necessary (your original code used int(-1) which is also redundant because -1 is already an int). If it’s a number already, then this whole if statement is unnecessary.

In general, the in operator is the right way to figure out if a thing is in another thing; only use methods like find or index if you need the actual index. The any function combined with a generator expression (for ... in ...) lets you apply that in check a bunch of times to a bunch of different items instead of needing to spell out each individual one.

If this is an XY problem and you’re really trying to figure out if num[p] can or can’t be interpreted as a float, you should instead do:

try:
    n = float(num[p])
    # do whatever
except ValueError:
    # do whatever else