Re: strange behavior of math.sqrt() in new 3.0 version
Tim Roberts wrote: Scott David Daniels scott.dani...@acm.org wrote: I avoid using single-letter variables except where I know the types from the name (so I use i, j, k, l, m, n as integers, s as string, and w, x, y, and z I am a little looser with (but usually float or complex). It's amazing to me that Fortran continues to live on in the hearts and minds of today's programmers. Fortran is alive and well. If you can find a Fortran 2003 compiler, you can even use classes. -- http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
Tim Roberts wrote: Scott David Daniels scott.dani...@acm.org wrote: I avoid using single-letter variables except where I know the types from the name (so I use i, j, k, l, m, n as integers, s as string, and w, x, y, and z I am a little looser with (but usually float or complex). It's amazing to me that Fortran continues to live on in the hearts and minds of today's programmers. Well I think it's more that the original Fortran programmers were applied mathematicians, who have always tended to use i through m as integer variables, and that usage continues both in programming and mathematics today. regards Steve -- Steve Holden+1 571 484 6266 +1 800 494 3119 Holden Web LLC http://www.holdenweb.com/ -- http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
Scott David Daniels scott.dani...@acm.org wrote: I avoid using single-letter variables except where I know the types from the name (so I use i, j, k, l, m, n as integers, s as string, and w, x, y, and z I am a little looser with (but usually float or complex). It's amazing to me that Fortran continues to live on in the hearts and minds of today's programmers. -- Tim Roberts, t...@probo.com Providenza Boekelheide, Inc. -- http://mail.python.org/mailman/listinfo/python-list
strange behavior of math.sqrt() in new 3.0 version
I'm a newbee trying 3.0 Please help with math.sqrt() At the command line this function works correctly import math n = input(enter a number ) s = math.sqrt(n) An entry of 9 or 9.0 will yield 3.0 Yet the same code in a script gives an error message Script1 import math n = input(enter a number ) s = math.sqrt(n) Traceback (most recent call last) : File stdin, line 1, in module File script1.py line 3 in module s = math.sqrt(n) TypeError : a float is required Entering 9 or 9.0 gives same error message. According to the math module the results of all functions are floats. However it says nothing about inputs. Strangely the above code runs fine in version 2.5 ( ? ) and will handle large integers. I've read the documentation for 3.0 including the section Floating Point Arithmetic: Issues Limitations and it helps nada. -- http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
David Lemper wrote:
I'm a newbee trying 3.0 Please help with math.sqrt()
At the command line this function works correctly
import math
n = input(enter a number )
s = math.sqrt(n)
An entry of 9 or 9.0 will yield 3.0
Yet the same code in a script gives an error message
Script1
import math
n = input(enter a number )
s = math.sqrt(n)
... TypeError : a float is required
Strangely the above code runs fine in version 2.5 ( ? ) ...
OK, here's what's going on:
at the command line in 2.X, the builtin function input reads a string
and returns the eval of that string. This is a bit of a safety issue.
I suspect when it worked from the command line, you were using a 2.X
command line inadvertently. in 2.X, you'll get similar errors if you
use raw_input instead of input.
The input function in 3.0 is the same as the raw_input function
in 2.X. I would suggest using:
import math
value = float(input(enter a number ))
root = math.sqrt(value)
print('root(%s) == %s' % (value, root))
I avoid using single-letter variables except where I know the types
from the name (so I use i, j, k, l, m, n as integers, s as string,
and w, x, y, and z I am a little looser with (but usually float or
complex).
--Scott David Daniels
scott.dani...@acm.org
--
http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
On Fri, Dec 26, 2008 at 1:52 PM, da...@bag.python.org wrote: I'm a newbee trying 3.0 Please help with math.sqrt() At the command line this function works correctly import math n = input(enter a number ) raw_input() was renamed input() in Python 3.0, and it returns a *string*, not a *number*. Therefore, you need to convert the string to an int or float. Change the line to: n = float(input(enter a number )) And it should work just fine. s = math.sqrt(n) An entry of 9 or 9.0 will yield 3.0 Yet the same code in a script gives an error message Script1 import math n = input(enter a number ) s = math.sqrt(n) Traceback (most recent call last) : File stdin, line 1, in module File script1.py line 3 in module s = math.sqrt(n) TypeError : a float is required You're currently giving it a string, not a number, which is nonsensical, hence the TypeError. I presume ints would be coerced to floats by the function. Entering 9 or 9.0 gives same error message. According to the math module the results of all functions are floats. However it says nothing about inputs. Strangely the above code runs fine in version 2.5 ( ? ) and will handle large integers. I've read the documentation for 3.0 including the section Floating Point Arithmetic: Issues Limitations and it helps nada. You should read What's New in Python 3.0, it covers changes such as the one you've encountered - http://docs.python.org/3.0/whatsnew/3.0.html Cheers, Chris -- Follow the path of the Iguana... http://rebertia.com -- http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
En Fri, 26 Dec 2008 19:52:24 -0200, da...@bag.python.org escribió: I'm a newbee trying 3.0 Please help with math.sqrt() At the command line this function works correctly import math n = input(enter a number ) s = math.sqrt(n) An entry of 9 or 9.0 will yield 3.0 Yet the same code in a script gives an error message Script1 import math n = input(enter a number ) s = math.sqrt(n) Traceback (most recent call last) : File stdin, line 1, in module File script1.py line 3 in module s = math.sqrt(n) TypeError : a float is required Entering 9 or 9.0 gives same error message. According to the math module the results of all functions are floats. However it says nothing about inputs. Strangely the above code runs fine in version 2.5 ( ? ) and will handle large integers. I've read the documentation for 3.0 including the section Floating Point Arithmetic: Issues Limitations and it helps nada. And you won't find nothing - the change is in input behavior, not in the math functions. For versions prior to 3.0, there are: raw_input(message) - string typed input(message) - result of evaluating the string typed raw_input just returns whatever you type, as a string. Using the input function, Python evaluates whatever you type to obtain a result: if you type the three characters nine dot zero the result is the double 9.0; you can even type (17+1)/2.0 to get the same value (try it with your Python 2.5) Since version 3.0, input behaves as raw_input in the older versions, and there is no builtin function equivalent to the old input function. Use this instead: n = float(input(enter a number )) -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: strange behavior of math.sqrt() in new 3.0 version
On Dec 27, 8:52 am, David Lemper wrote: I'm a newbee trying 3.0 Please help with math.sqrt() math.sqrt() is not the problem. At the command line this function works correctly import math n = input(enter a number ) s = math.sqrt(n) An entry of 9 or 9.0 will yield 3.0 I can't reproduce this. See below. Are you sure that you weren't using Python 2.x? Start with a fresh Python 3.0, and please copy/paste *exactly* what is on the screen, like I've done: C:\junk\python30\python Python 3.0 (r30:67507, Dec 3 2008, 20:14:27) [MSC v.1500 32 bit (Intel)] on win32 Type help, copyright, credits or license for more information. n = input(float- ) float- 9 type(n) class 'str' repr(n) '9' import math s = math.sqrt(n) Traceback (most recent call last): File stdin, line 1, in module TypeError: a float is required All of the above is exactly as expected. input() returns a string. math.sqrt expects a float. Yet the same code in a script gives an error message Script1 import math n = input(enter a number ) s = math.sqrt(n) Traceback (most recent call last) : File stdin, line 1, in module File script1.py line 3 in module s = math.sqrt(n) TypeError : a float is required Entering 9 or 9.0 gives same error message. Again, as expected. According to the math module the results of all functions are floats. However it says nothing about inputs. Strangely the above code runs fine in version 2.5 ( ? ) and will handle large integers. Suggestion: either choose ONE of 2.6 and 3.0 to learn Python, or plan to spend a lot of time being confused, or reading What's new in Python 3.0 -- a relevant snippet of which is: PEP 3111: raw_input() was renamed to input(). That is, the new input() function reads a line from sys.stdin and returns it with the trailing newline stripped. It raises EOFError if the input is terminated prematurely. To get the old behavior of input(), use eval(input()). HTH, John -- http://mail.python.org/mailman/listinfo/python-list
