value error in my code (python 3.0) 'chr() arg not in range'
I have to take in a string and an integer value and check if the string is lowercase or uppercase, and based on that I have to increment it by number k. for eg if k=4 and string is 'ABab' it should give the output 'EFef'.
This is my code only for checking lowercase. Unfortunately is giving ValueError.
s=input()
k=int(input())
l=
for i in s:
if i.islower():
if 97>=(ord(i)+k)<=122:
l.append(chr(ord(i)+k))
else:
k=k-122
if 97>=(ord(i)+k)<=122:
l.append((chr(ord(i)+k)))
break
else:
continue
print(l)
python string
edited Nov 22 '18 at 17:00
James Z
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
add a comment |
I have to take in a string and an integer value and check if the string is lowercase or uppercase, and based on that I have to increment it by number k. for eg if k=4 and string is 'ABab' it should give the output 'EFef'.
This is my code only for checking lowercase. Unfortunately is giving ValueError.
s=input()
k=int(input())
l=
for i in s:
if i.islower():
if 97>=(ord(i)+k)<=122:
l.append(chr(ord(i)+k))
else:
k=k-122
if 97>=(ord(i)+k)<=122:
l.append((chr(ord(i)+k)))
break
else:
continue
print(l)
python string
edited Nov 22 '18 at 17:00
James Z
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
add a comment |
I have to take in a string and an integer value and check if the string is lowercase or uppercase, and based on that I have to increment it by number k. for eg if k=4 and string is 'ABab' it should give the output 'EFef'.
This is my code only for checking lowercase. Unfortunately is giving ValueError.
s=input()
k=int(input())
l=
for i in s:
if i.islower():
if 97>=(ord(i)+k)<=122:
l.append(chr(ord(i)+k))
else:
k=k-122
if 97>=(ord(i)+k)<=122:
l.append((chr(ord(i)+k)))
break
else:
continue
print(l)
python string
edited Nov 22 '18 at 17:00
James Z
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
I have to take in a string and an integer value and check if the string is lowercase or uppercase, and based on that I have to increment it by number k. for eg if k=4 and string is 'ABab' it should give the output 'EFef'.
This is my code only for checking lowercase. Unfortunately is giving ValueError.
s=input()
k=int(input())
l=
for i in s:
if i.islower():
if 97>=(ord(i)+k)<=122:
l.append(chr(ord(i)+k))
else:
k=k-122
if 97>=(ord(i)+k)<=122:
l.append((chr(ord(i)+k)))
break
else:
continue
print(l)
python string
python string
edited Nov 22 '18 at 17:00
James Z
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
edited Nov 22 '18 at 17:00
James Z
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
edited Nov 22 '18 at 17:00
James Z
11.1k71835
edited Nov 22 '18 at 17:00
James Z
11.1k71835
edited Nov 22 '18 at 17:00
James Z
11.1k71835
11.1k71835
asked Nov 22 '18 at 15:05
palakpalak
94
asked Nov 22 '18 at 15:05
palakpalak
94
asked Nov 22 '18 at 15:05
palakpalak
94
94
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The traceback shows where the error occurs.
Traceback (most recent call last):
File "C:/Users/rob/test.py", line 11, in <module>
l.append((chr(ord(i)+k)))
ValueError: chr() arg not in range(0x110000)
You are passing an argument to chr that is not within the allowed range. As described here:
The valid range for the argument is from 0 through 1,114,111 (0x10FFFF
in base 16). ValueError will be raised if i is outside that range.
This is because you have changed the value of k to (probably) be a large negative number:
k=k-122
So the result of ord(i)+k is also often negative. Negative numbers are not in the allowed range, so the call to chr fails.
There are lots of other problems with your code, and I don't think you'd learn much if I just wrote "my solution" to the problem. Another thing you might want to look at to begin with is that:
if 97>=(ord(i)+k)<=122:
doesn't do what you want, you probably want:
if 97<=(ord(i)+k)<=122:
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of theprint(l)at the end of the question.
– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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active
oldest
votes
The traceback shows where the error occurs.
Traceback (most recent call last):
File "C:/Users/rob/test.py", line 11, in <module>
l.append((chr(ord(i)+k)))
ValueError: chr() arg not in range(0x110000)
You are passing an argument to chr that is not within the allowed range. As described here:
The valid range for the argument is from 0 through 1,114,111 (0x10FFFF
in base 16). ValueError will be raised if i is outside that range.
This is because you have changed the value of k to (probably) be a large negative number:
k=k-122
So the result of ord(i)+k is also often negative. Negative numbers are not in the allowed range, so the call to chr fails.
There are lots of other problems with your code, and I don't think you'd learn much if I just wrote "my solution" to the problem. Another thing you might want to look at to begin with is that:
if 97>=(ord(i)+k)<=122:
doesn't do what you want, you probably want:
if 97<=(ord(i)+k)<=122:
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of theprint(l)at the end of the question.
– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
add a comment |
The traceback shows where the error occurs.
Traceback (most recent call last):
File "C:/Users/rob/test.py", line 11, in <module>
l.append((chr(ord(i)+k)))
ValueError: chr() arg not in range(0x110000)
You are passing an argument to chr that is not within the allowed range. As described here:
The valid range for the argument is from 0 through 1,114,111 (0x10FFFF
in base 16). ValueError will be raised if i is outside that range.
This is because you have changed the value of k to (probably) be a large negative number:
k=k-122
So the result of ord(i)+k is also often negative. Negative numbers are not in the allowed range, so the call to chr fails.
There are lots of other problems with your code, and I don't think you'd learn much if I just wrote "my solution" to the problem. Another thing you might want to look at to begin with is that:
if 97>=(ord(i)+k)<=122:
doesn't do what you want, you probably want:
if 97<=(ord(i)+k)<=122:
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of theprint(l)at the end of the question.
– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
add a comment |
The traceback shows where the error occurs.
Traceback (most recent call last):
File "C:/Users/rob/test.py", line 11, in <module>
l.append((chr(ord(i)+k)))
ValueError: chr() arg not in range(0x110000)
You are passing an argument to chr that is not within the allowed range. As described here:
The valid range for the argument is from 0 through 1,114,111 (0x10FFFF
in base 16). ValueError will be raised if i is outside that range.
This is because you have changed the value of k to (probably) be a large negative number:
k=k-122
So the result of ord(i)+k is also often negative. Negative numbers are not in the allowed range, so the call to chr fails.
There are lots of other problems with your code, and I don't think you'd learn much if I just wrote "my solution" to the problem. Another thing you might want to look at to begin with is that:
if 97>=(ord(i)+k)<=122:
doesn't do what you want, you probably want:
if 97<=(ord(i)+k)<=122:
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
The traceback shows where the error occurs.
Traceback (most recent call last):
File "C:/Users/rob/test.py", line 11, in <module>
l.append((chr(ord(i)+k)))
ValueError: chr() arg not in range(0x110000)
You are passing an argument to chr that is not within the allowed range. As described here:
The valid range for the argument is from 0 through 1,114,111 (0x10FFFF
in base 16). ValueError will be raised if i is outside that range.
This is because you have changed the value of k to (probably) be a large negative number:
k=k-122
So the result of ord(i)+k is also often negative. Negative numbers are not in the allowed range, so the call to chr fails.
There are lots of other problems with your code, and I don't think you'd learn much if I just wrote "my solution" to the problem. Another thing you might want to look at to begin with is that:
if 97>=(ord(i)+k)<=122:
doesn't do what you want, you probably want:
if 97<=(ord(i)+k)<=122:
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
edited Nov 22 '18 at 15:40
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
answered Nov 22 '18 at 15:24
Rob BrichenoRob Bricheno
2,325218
2,325218
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of theprint(l)at the end of the question.
– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
add a comment |
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of theprint(l)at the end of the question.
– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
1
1
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
Please note that the range "0-255" is a Python 2 thing only - nowadays most people use Python 3, and CHR and ORD give results in unicode code points. That said, the OP is likely using Python 2, which should be noted in the answer, them.
– jsbueno
Nov 22 '18 at 15:35
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of the
print(l) at the end of the question.– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno Thanks, good catch, I've updated my answer for Python 3. I'm sure OP is using Python 3 because of the
print(l) at the end of the question.– Rob Bricheno
Nov 22 '18 at 15:42
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
@jsbueno I think it's more accurate to say that most people learning Python are using Python 3 (or should be, anyway). The installed base of Python 2 code is still quite large (though probably larger than it should be, with Python 2 support coming to an end).
– chepner
Nov 22 '18 at 15:47
add a comment |
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