想在python中搜索给定字符串中的数值

Question

string3 = "abc 123 $$%%"

list1 = string3.split()
print(list1)
for i in list1:
    if int(i) > 0:
        print("it's a number")
    else:
        print("not a number")

Getting below error :

if int(i) > 0:
ValueError: invalid literal for int() with base 10: 'abc'

Answer 1

>>> str = "abc 123 $$%%"
>>> [int(s) for s in str.split() if s.isdigit()]
[123]

Answer 2

use i.isdigit()

string3 = "abc 123 $$%%"

list1 = string3.split() 
print(list1)
for i in list1:
    if i.isdigit():
        print("it's a number") 
    else: 
        print("not a number")

Answer 3

Fancy way:

>>> s = "abc 123 $$%%"
>>> map(int,filter(str.isdigit,s.split()))
[123]

Explanation:

• s.split() is splitting the string on spaces and generates: ['abc', '123', '$$%%']
• str.isdigit is a function which returns True if all characters in the argument are digits.
• filter filters out elements of a list which do not pass the test. First argument is the test function: str.isdigit , second argument is the list.
• Finally, map transforms one list to another. First argument is the transform function int , second argument is the list found from filter .

Answer 4

try this

string3 = "abc 123 $$%%"

list1 = string3.split()
print(list1)
for i in list1:
    if i.isdigit():
        print("it's a number")
    else:
        print("not a number")

Output :
['abc', '123', '$$%%']
not a number
it's a number
not a number

Answer 5

string3 = "abc 123 $$%%"

list1 = string3.split()
print(list1)
for i in list1:
    try:
        int(i)
        print("It is a number")
    except ValueError:
        print("It is not a number")

Try this code