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basic if stuff- testing ranges

P: n/a
Sheesh, I've been going spare trying to find how to do this short-hand:
if 0 x < 20: print "within"

So that x must be 0 and < 20.

I usually do:
if x 0 and x < 20: print "within"

What's the rule? Does it even exist?
I read something like it recently on the list but can't find it, that's
where I got the urge to try it from. I can't find anything in the docs, but
then again (imho) the Python docs are like a tangled jungle...
\d

Nov 25 '07 #1
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P: n/a
Mel
Donn Ingle wrote:
Sheesh, I've been going spare trying to find how to do this short-hand:
if 0 x < 20: print "within"

So that x must be 0 and < 20.

I usually do:
if x 0 and x < 20: print "within"

What's the rule? Does it even exist?
if 0 < x < 20:
?
Mel.
Nov 25 '07 #2

P: n/a
Donn Ingle a écrit :
Sheesh, I've been going spare trying to find how to do this short-hand:
if 0 x < 20: print "within"
you mean : 0 < x < 20 ?

or

x in xrange(1,20) ?
>
So that x must be 0 and < 20.

I usually do:
if x 0 and x < 20: print "within"

What's the rule? Does it even exist?
is that hard to check it at the repl ?
I read something like it recently on the list but can't find it, that's
where I got the urge to try it from. I can't find anything in the docs, but
then again (imho) the Python docs are like a tangled jungle...
<something smug, refrained/>
>

\d
Nov 25 '07 #3

P: n/a
On Nov 26, 5:49 am, Donn Ingle <donn.in...@gmail.comwrote:
Sheesh, I've been going spare trying to find how to do this short-hand:
if 0 x < 20: print "within"
That means "if x LESS THAN 0 and x < 20".
>
So that x must be 0 and < 20.
So try
if 0 < x < 20:
>
I usually do:
if x 0 and x < 20: print "within"

What's the rule? Does it even exist?
I read something like it recently on the list but can't find it, that's
where I got the urge to try it from. I can't find anything in the docs, but
then again (imho) the Python docs are like a tangled jungle...
Likely manuals: Tutorial & Reference
Tutorial: check contents, "if statement" looks possible, but no luck
Reference: check contents, "comparisons" looks possible, and
http://docs.python.org/ref/comparisons.html says:
"""
Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent
to x < y and y <= z, except that y is evaluated only once (but in both
cases z is not evaluated at all when x < y is found to be false).
"""
Nov 25 '07 #4

P: n/a
you mean : 0 < x < 20 ?
Yes. I had gotten the impression that there was some Python form of:
if NUMBER test VAR test NUMBER:

Part of the question was to discover if I was smoking my socks :)
x in range(1,20) ?
Sure, that's okay, but it has clarity issues, and is calling a func.
>but then again (imho) the Python docs are like a tangled jungle...
<something smug, refrained/>
Well, I said it was MHO and if it was easier (for me) to find answers in the
docs I'd have an easier time of it.

\d

Nov 26 '07 #5

P: n/a
Mel wrote:
if 0 < x < 20:
?
I take it I was tripping then. That's okay, it seemed a little too weird
anyway :)

\d

Nov 26 '07 #6

P: n/a
>if 0 x < 20: print "within"
That means "if x LESS THAN 0 and x < 20".
Oh, bugger. It's tricky.
So try
if 0 < x < 20:
Thanks. I was flipping signs in my tests, but I guess I flipped both and got
myself all confused.
Likely manuals: Tutorial & Reference
Tutorial: check contents, "if statement" looks possible, but no luck
Yes, I got that far.
Reference: check contents, "comparisons" looks possible, and
Thanks again. I find the reference is laid-out in a way that I don't find
intuitive and every time I look for something I fail. I even grep through
the folder to get a clue, which shows how poor the index is (to me)!

Many thanks for the help!
\d

Nov 26 '07 #7

P: n/a
On Nov 25, 6:49 pm, Donn Ingle <donn.in...@gmail.comwrote:
Sheesh, I've been going spare trying to find how to do this short-hand:
if 0 x < 20: print "within"

So that x must be 0 and < 20.

I usually do:
if x 0 and x < 20: print "within"

What's the rule? Does it even exist?
I read something like it recently on the list but can't find it, that's
where I got the urge to try it from. I can't find anything in the docs, but
then again (imho) the Python docs are like a tangled jungle...

\d
The output of the following program might help:

# chained_comparisons.py
complist = '< <= == != >= >'.split()
for lhs in complist:
for rhs in complist:
print "\n1 %2s x %2s 3:" % (lhs, rhs)
for x in range(5):
chain = " 1 %2s %i %2s 3" % (lhs, x, rhs)
print chain," is ", eval(chain)
- Paddy.
Nov 26 '07 #8

P: n/a
Donn Ingle wrote:
>x in range(1,20) ?
Sure, that's okay, but it has clarity issues, and is calling a func.
and it requires that x is integral (1.0 is in the range, 1.001 is not),
and becomes dog slow when the range gets larger. Not a good idea.

Peter
Nov 26 '07 #9

P: n/a
The output of the following program might help:
Hey, nifty! Thanks Paddy.

\d

Nov 26 '07 #10

P: n/a

On Nov 25, 2007, at 9:49 PM, Donn Ingle wrote:
>>if 0 x < 20: print "within"
That means "if x LESS THAN 0 and x < 20".
Oh, bugger. It's tricky.
>So try
if 0 < x < 20:
Thanks. I was flipping signs in my tests, but I guess I flipped
both and got
myself all confused.
>Likely manuals: Tutorial & Reference
Tutorial: check contents, "if statement" looks possible, but no luck
Yes, I got that far.
>Reference: check contents, "comparisons" looks possible, and
Thanks again. I find the reference is laid-out in a way that I
don't find
intuitive and every time I look for something I fail. I even grep
through
the folder to get a clue, which shows how poor the index is (to me)!
Then use one of the quick references here: http://rgruet.free.fr/.

Erik Jones

Software Developer | Emma®
er**@myemma.com
800.595.4401 or 615.292.5888
615.292.0777 (fax)

Emma helps organizations everywhere communicate & market in style.
Visit us online at http://www.myemma.com
Nov 26 '07 #11

P: n/a
On Nov 26, 2007, at 2:29 AM, Peter Otten wrote:
Donn Ingle wrote:
>>x in range(1,20) ?
Sure, that's okay, but it has clarity issues, and is calling a func.

and it requires that x is integral (1.0 is in the range, 1.001 is
not),
and becomes dog slow when the range gets larger. Not a good idea.
That is because range() is not a range in the abstract sense (i.e.
simply defining bounds that can be tested for set membership) but are
used to create lists (or, in the case of xrange(), successive values)
between the bounds given in the params. So, saying x in range(1,20)
is not asking if x is between 1 and 20 but, rather, if x is a member
of the values genereated by the range function with params 1 and 20.
So, yes, using range()

Erik Jones

Software Developer | Emma®
er**@myemma.com
800.595.4401 or 615.292.5888
615.292.0777 (fax)

Emma helps organizations everywhere communicate & market in style.
Visit us online at http://www.myemma.com
Nov 26 '07 #12

P: n/a
On Nov 27, 3:01 am, Erik Jones <e...@myemma.comwrote:
On Nov 25, 2007, at 9:49 PM, Donn Ingle wrote:
>if 0 x < 20: print "within"
That means "if x LESS THAN 0 and x < 20".
Oh, bugger. It's tricky.
So try
if 0 < x < 20:
Thanks. I was flipping signs in my tests, but I guess I flipped
both and got
myself all confused.
Likely manuals: Tutorial & Reference
Tutorial: check contents, "if statement" looks possible, but no luck
Yes, I got that far.
Reference: check contents, "comparisons" looks possible, and
Thanks again. I find the reference is laid-out in a way that I
don't find
intuitive and every time I look for something I fail. I even grep
through
the folder to get a clue, which shows how poor the index is (to me)!

Then use one of the quick references here:http://rgruet.free.fr/.
Generally excellent references, but "X < Y < Z < W has expected
meaning, unlike C" is not much help to people who have not been
exposed to similar notation (e.g. 0 <= angle < 2 * pi) in other
disciplines and/or know C merely as a member of the same set as X, Y,
Z and W.
Nov 26 '07 #13

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