I am trying to check if in a string(eg "Hello my name is 3") if there is a number. In my example the code I want to write would identify the number 3.
I have tried the code bellow. I know the "int(range(1,10)): " part is not going to work at all but I put it there to show what im trying to achieve if that makes sense.
for x in text:
if x != " " or x != int(range(1,10)):
print(x)
To test if string contains a number:
>>> def hasNumbers(inputString):
... return any(char.isdigit() for char in inputString)
>>> hasNumbers("Hello my name is 3")
True
>>> hasNumbers("Hello my name is p")
False
To extract number:
>>> int(filter(str.isdigit, inputString))
3
You've got the right idea, you just need another for loop after the first.
for i in range(1,10): #this will only check 1-9
if i != int(x):
print(x)
Alternatively, you could use regular expressions, which would be faster with a larger range of numbers;
import re
if not re.match("[0-9]",x):
print(x)
This checks for a digit 0-9 in x, and prints if it's not there.
If you want to just test that your string has a proper decimal, try this
test='Hello my name is 3'
if any((x.isdecimal() for x in test.split())):
# Yes
Note: This won' work for example, Hello my name is3
If you want to check digit-by-digit, then just remove the split.
any((x.isdecimal() for x in test))
Use regular expression library: regex
import re
s = "Hello my name is 3"
re.findall("\d*", s)
3
