I have a select statement which return capacity as exponential value e.g.
Capacity=5.4835615662E+003
in Perl code
I am using a db2 database, and if I explicitly run a query in database it returns
5483.5615662
but when I use next select query when I use capacity value in condition it doesn't match
e.g. pseudo code is as below,
my $capacity = 'SELECT capacity FROM table';
# it returns $capacity = 5.4835615662E+003
my $result = "SELECT MEASUREMENT FROM TABLE WHERE CAPACITY = $capacity";
Here $capacity is 5.4835615662E+003, so it does not match any row in the table. It should be 5483.5615662.
How to convert exponential value to float without rounding off?
You are interpolating the value of $capacity into a string. Instead, you should use placeholders as in:
my $sth = $dbh->prepare(q{SELECT MEASUREMENT FROM TABLE WHERE CAPACITY=?});
$sth->execute($capacity);
It is hard to say if there are any other problems because the code snippets you provide don't really do anything.
It is likely that the number stored in the database is not exactly 5483.5615662 and that is just the displayed string when you query it.
If possible, I would recommend taking #Сухой27's advice and letting the database do the work for you:
SELECT MEASUREMENT FROM TABLE
WHERE CAPACITY = (SELECT CAPACITY FROM TABLE where ..?)
Alternatively, decide ahead of time how many digits past the decimal point really matter and use ROUND or similar functionality:
my $sth = $dbh->prepare(q{
SELECT MEASUREMENT FROM TABLE
WHERE ROUND(CAPACITY, 6)=ROUND(?, 6)
});
$sth->execute($capacity);
Please take a look at Why doesn't this sql query return any results comparing floating point numbers?
I'm concerned about the 5.4835615662+003 that you show in your question. That isn't a valid representation of a number, and it means just 5.4835615662 + 3. You need an E or an e before the exponent to use it as it is
There is also an issue with comparing floating-point values, whereby two numbers that are essentially equal may have a slightly different binary representation, and so will not compare as equal. If your value has been converted to a string (and that seems highly likely, as Perl will not use an exponent to display 5483.5615662 unless told to do so) and back again to floating point, then it is extremely unlikely to result in exactly the same value. Your comparisons will always fail
In Perl, and most other languages, a numeric values has no specific format. For example, if I run this
perl -E 'say 5.4835615662E+003'
I get the output
5483.5615662
showing that the two string representations are equivalent
It would help to see exactly how you got the value of $capacity from the database, because if it were a simple number then it wouldn't use the scientific representation. You would have to use sprintf to get what you have shown
SQL is the same and doesn't care about the format of the number as long as it's valid, so if you wrote
SELECT measurement FROM table WHERE capacity = 5.4835615662E+003
then you would get a result where capacity is exactly equal to that value. But since it has been trimmed to eleven significant digits, you are hugely unlikely to find the record that the value came from, unless it contains 5483.56156620000000000
Update
If I run
perl -MMath::Trig=pi -E 'for (0 .. 20) { $x = pi * 10**$_; say qq{$x}; }'
I get this result
3.14159265358979
31.4159265358979
314.159265358979
3141.59265358979
31415.9265358979
314159.265358979
3141592.65358979
31415926.5358979
314159265.358979
3141592653.58979
31415926535.8979
314159265358.979
3141592653589.79
31415926535897.9
314159265358979
3.14159265358979e+015
3.14159265358979e+016
3.14159265358979e+017
3.14159265358979e+018
3.14159265358979e+019
3.14159265358979e+020
So by default Perl won't resort to using scientific notation until the value reaches 1015. It clearly doesn't apply to 5483.5615662. Something has coerced the floating-point value in the question to a much less precise string in scientific notation. Comparing that for equality doesn't stand a chance of succeeding
So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?
This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.
The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);
I have a few numbers in a file in a variety of formats: 8.3, 0.001, 9e-18. I'm looking for an easy way to read them in and store them without any loss of precision. This would be easy in AWK, but how's it done in Perl? I'm only open to using Perl. Thanks!
Also, I was wondering if there's an easy way to print them in an appropriate format. For example, 8.3 should be printed as "8.3" not "8.3e0"
If they're text strings, then reading them into Perl as strings and writing them back out as strings shouldn't result in any loss of precision. If you have to do arithmetic on them, then I suggest installing the CPAN module Math::BigFloat to ensure that you don't lose any precision to rounding.
As to your second question, Perl doesn't do any reformatting unless you ask it to:
$ perl -le 'print 8.3'
8.3
Am I missing something?
From http://perldoc.perl.org/perlnumber.html:
Perl can internally represent numbers in 3 different ways: as native
integers, as native floating point numbers, and as decimal strings.
Decimal strings may have an exponential notation part, as in
"12.34e-56" . Native here means "a format supported by the C compiler
which was used to build perl".
This means that printing the number out depends on how the number is stored internal to perl, which means, in turn, that you have to know how the number is represented on input.
By and large, Perl will just do the right thing, but you should know how what compiler was used, how it represents numbers internally, and how to print those numbers. For example:
$ perldoc -f int
int EXPR
int Returns the integer portion of EXPR. If EXPR is omitted, uses $_. You should
not use this function for rounding: one because it truncates towards 0, and two
because machine representations of floating-point numbers can sometimes produce
counterintuitive results. For example, "int(-6.725/0.025)" produces -268 rather than
the correct -269; that's because it's really more like -268.99999999999994315658
instead. Usually, the "sprintf", "printf", or the "POSIX::floor" and
"POSIX::ceil" functions will serve you better than will int().
I think that if you want to read a number in explicitly as a string, your best bet would be to use unpack() with the 'A*' format.
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Closed 12 years ago.
Possible Duplicate:
How do I fix this Perl code so that 1.1 + 2.2 == 3.3?
I'm working on a Perl script that compares strings representing gene models and prints out a summary of the comparison. If the gene models match perfectly, I print out a very terse summary, but if they are different, the summary is quite verbose.
The script looks at the value of a variable to determine whether it should do the terse or verbose summary--if the variable is equal to 1, it should print the terse summary; otherwise, it should print the verbose summary.
Since the value is numeric (a float), I've been using the == operator to do the comparison.
if($stats->{overall_simple_matching_coefficient} == 1)
{
print "Gene structures match perfectly!\n";
}
This worked correctly for all of my tests and even for most of the new cases I am running now, but I found a weird case where the value was equal to 1 but the above comparison failed. I have not been able to figure out why the comparison failed, and stranger yet, when I changed the == operator to the eq operator, it seemed to work fine.
I thought the == was for numerical comparison and eq was for string comparison. Am I missing something here?
Update: If I print out the value right before the comparison...
printf("Test: '%f', '%d', '%s'\n", $stats->{overall_simple_matching_coefficient}, $stats->{overall_simple_matching_coefficient}, $stats->{overall_simple_matching_coefficient});
...I get this.
Test: '1.000000', '0', '1'
The first thing any computer language teacher should teach you about any computer language is that YOU CANNOT COMPARE FLOATS FOR EQUALITY. This is true of any language. Floating point arithmetic is not exact, and two floats that look like they're the same will be different in the insignificant digits somewhere where you can't see it. Instead, you can only compare that they are close to each other - like
if (abs(stats->{overall_simple_matching_coefficient)-1) < 0.0001)
What do you get if you print the value of $stats->{overall_simple_matching_coefficient} just before the comparison? If it's 1, try printf with a format of "%20.10f". I strongly suspect you have some rounding error (less then 1e-6) accumulated in the variable and it's not comparing equal numerically. However when converted to string, since the error is right of the 6th decimal place, and the default string format is to six places, it compares equal.