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Views: 504 | Replies: 4
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Join Date: Oct 2007
Posts: 2
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Newbie Poster
Python bug??? Or just a stupid question? Hello,
I am a newbie in Python and maybe this is a stupid question, but when I run this code I receive a very strange output.
At first I supposed that I will receive something like this:
>>>
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
>>>
But finally I was really surprised. Output looks like this:
>>>
0.1
0.2
0.4
0.5
0.6
0.7
>>>
Could somebody tells me what I am doing wrong?
I am a newbie in Python and maybe this is a stupid question, but when I run this code I receive a very strange output.
status = 0
for i in range(10):
status = status + 0.1
if status == 0.1:
print status
elif status == 0.2:
print status
elif status == 0.3:
status
elif status == 0.4:
print status
elif status == 0.5:
print status
elif status == 0.6:
print status
elif status == 0.7:
print status
elif status == 0.8:
print status
elif status == 0.9:
print status
elif status == 1.0:
print status>>>
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
>>>
But finally I was really surprised. Output looks like this:
>>>
0.1
0.2
0.4
0.5
0.6
0.7
>>>
Could somebody tells me what I am doing wrong?
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Join Date: Oct 2004
Posts: 2,529
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Solved Threads: 178
This happens with many computer languages. You are trying to compare floating point values directly. Your 0.8 might be more like 0.799999999999999 and the compare flunks!
You can make this work by giving it a range you can compare within ...
You can make this work by giving it a range you can compare within ...
if 0.79999 < status < 0.80001:
... do something` Last edited by vegaseat : Oct 5th, 2007 at 10:22 pm.
May 'the Google' be with you!
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Join Date: Dec 2006
Posts: 468
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Posting Pro in Training
Look at the line following
elif status == 0.3:
And why do you have if/elif? Why not just print the value?
elif status == 0.3:
And why do you have if/elif? Why not just print the value?
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Join Date: Oct 2007
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Newbie Poster
Thanks for an answer. It was just an example which really surprise me. But is there a way how to compare two float numbers without unpredictable results?
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Join Date: Jul 2006
Posts: 562
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Posting Pro
Well, there are actually three issues in the code:
(1) is a matter of efficiency, which woooee already raised.
does what you wanted to do, and is cleaner to read. One payoff of this approach is that...
(2) ...in your original code, you didn't get 0.3 in the output because you left out the print statement. The lines
have the effect of evaluating, but not printing, status. And then throwing away the result, because you aren't doing anything with it.
(3) Finally, comparing floating points is a common source of bugs. Even if we worked with BCD, so that we didn't have to work about base-2 to base-10 errors, still and all, our decimal approximations would still lead to imprecision. For example, the test π == 3.14159265358979 is False.
Vega gives a very good approach to this by selecting a range. My own approach is to define a function that is in my code base:
This function returns True if a is within the "halo" of b; that is, if b-epsilon < a < b+epsilon.
You can then use it like this:
Hope that helps,
Jeff
(1) is a matter of efficiency, which woooee already raised.
Python Syntax (Toggle Plain Text)
does what you wanted to do, and is cleaner to read. One payoff of this approach is that...
(2) ...in your original code, you didn't get 0.3 in the output because you left out the print statement. The lines
Python Syntax (Toggle Plain Text)
have the effect of evaluating, but not printing, status. And then throwing away the result, because you aren't doing anything with it.
(3) Finally, comparing floating points is a common source of bugs. Even if we worked with BCD, so that we didn't have to work about base-2 to base-10 errors, still and all, our decimal approximations would still lead to imprecision. For example, the test π == 3.14159265358979 is False.
Vega gives a very good approach to this by selecting a range. My own approach is to define a function that is in my code base:
Python Syntax (Toggle Plain Text)
This function returns True if a is within the "halo" of b; that is, if b-epsilon < a < b+epsilon.
You can then use it like this:
Python Syntax (Toggle Plain Text)
Hope that helps,
Jeff
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