I'm having trouble with creating a function that takes an argument of a string and returns just the numbers in type str. My code looks something like this.:
def check_if_number(number_string):
for ch in number_string:
if ch.isdigit():
num = number_string[ch]
return num
print(check_if_number('1655t'), end='')
It should print: 1655
You should add each char to a string if it is a digit :
def check_if_number(number_string):
digit_string = ""
for ch in number_string:
if ch.isdigit():
digit_string += ch
return digit_string
print(check_if_number('1655t'))
# 1655
Note that you can also use the re library :
import re
re.sub("\D", "", "1655t")
#1655
This code replace every non-digit character (matched by \D) by an empty string.
Have a look at this question to find more way to do it
You can do this :
def check_if_number(number_string):
num = ""
for ix in number_string :
if ix.isdigit():
num += ix
return int(num)
This function will return 1655, if the input is 1655t. You return from the function once you scan the complete string and append all ints to a string.
ch which is confusing as all hell. This is terribly confusing for someone who is clearly a beginner.
ch is a character in the string, hence, you can't use it for indexing.The easy way would be just using filter and concatenating the filtered letters in the end:
def check_if_number(number_string):
return ''.join(filter(str.isdigit, number_string))
Assuming all numbers are together (i.e. No letters exist between digits like "1e234g7"),
def check_if_number(number_string):
num = ''
for ch in number_string:
if ch.isdigit():
num += ch
return num
print(check_if_number('1655t'), end='')
This will return a string that only has digits in it. If you want to return a type int, then simply do return int(num).
If you want to take floats into account as well:
def check_if_number(number_string):
num = ''
for ch in number_string:
if ch == '.' or ch.isdigit():
num += ch
return num
print(check_if_number('1655t'), end='')
Similarly, this will return a string. If you want type float, simply do return float(num).
I tried not to deviate too much from what you already had... I hope this helps.
I'm uncertain what check_leap_year() function is so I just commented it out and called the function you gave us instead.
When you pass anything into a function in order to manipulate it and spit it back out different than it came in you should create a new variable for the changed data set.
In this case I used num = ""
By initiating an empty string inside of the function I can now add the data that pass my condition to that new, empty string.
isdigit() returns a boolean. So instead of using if x.isdigit: get in the habit of checking the boolean. Try if x.isdigit() is True instead.
Return your finished new string AFTER it is finished. Make sure your return statement is indented correctly or you are only going to return one number instead of all of them. The return should be outside of the loop so that it will only return once the loop is done.
num_string = "1a2s3d4f5g666hhh777jjj888kkk9l0"
def check_if_number(number_string):
num = ""
for ch in number_string:
if ch.isdigit() is True:
num += ch
return num
print(check_if_number(num_string))
Output:
>>> 1234566677788890
isdigit() to determine if something is not a digit. In which case you would need to use the is operand. if x.isdigit() is False: A good rule of thumb is if you need it sometimes, but when you do not need it it gives you the same results, just make a habit of doing it. And if you ever have people reading your code in the future they will thank you
check_leap_yearfunction ? Also, what exact error you are getting ?numis just a single character,chis still a string you you need to convert that to anintafter checking, as you can't index with it. Are you trying to remove all non-digit characters or just check if the digit isn't only digits?