I just started studying pascal and I have to do a pascal program as homework.
I made it but I don't know how to count the number of decimal places in a real number (the number of digit after the ".").
I need it just to format well a real number (like write(real:0:dec) where dec is the number of decimal digit i don't know how to know). I'd like to do that because i don't want it in scientific notation or with many unnecessary zeros.
For example if a real number is 1.51 (x) and I write writeln(x:0:4); or WriteLn(Format('%*.*f', [0, 4, x])); it will show 1.5100 but I want it to be just 1.51; and if the number is like 1.513436, it will show only 1.5134 . So I would make it like writeln(x:0:dec); with something that makes dec the number of decimal digits of x.
The Format() function is normally used in situations like this.
WriteLn(Format('%*.*f', [0, dec, your_real_number]));
*.* is interpreted as total_characters.decimal_digits. Passing zero for the first means that width is adjusted according to how large your real is. The number of decimals can be a variable (dec), which you can adjust to your specification.
Update:
You mention that you want an exact representation of a float with respect to the number of decimals.
As mentioned in my comment, most floating point values does not have a finite number of decimals. And most decimal fractions cannot be represented by a binary type.
There are some libraries that can handle floating point values of arbitrary size. See TBigFloat for example. There is a formatting routine that can strip redundant zeroes from a decimal float.
Still, there is a possibility to remove trailing zeroes by using the general format specifier:
WriteLn(Format('%g',[your_real_number]));
You can specify the width and the number of significant digits as well.
For example, if you have input x=1.51 in real variable type, then you write only writeln(x), the output will be 1.5100000000. If you write writeln(x:0:3), the output will be 1.510 (3 digits after ".") ...
var x: real;
Begin
x:=1.51;
writeln(x); //output 1.5100000000
writeln(x:0:4); //output 1.5100 (4 digits after ".")
writeln(x:0:2); //output 1.51 (2 digits after ".")
readln;
End.
From your other example, if your input is 1.512426, with writeln(x:0:5) it will only show 5 digits after "." and the output will be 1.51242
var x: real;
Begin
x:=1.512426;
writeln(x); //output 1.5124260000
writeln(x:0:4); //output 1.5124 (4 digits after ".")
writeln(x:0:2); //output 1.51 (2 digits after ".")
readln;
End.
So, if you write writeln(x:0:dec) the output will be "dec" digits after "."
Hope this helps, I'm just trying to answer from a different perspective.
if you're actually doing a writeln() output, surely just
writeln(x);
would accomplish what you're after? If you actually want to count the number of decimals, you'd probably have to convert to a string, remove any trailing zeroes, and see where the decimal point landed.
Related
I would like to print a series of floats with varying amounts of numbers to the left of the decimal place. I would like these numbers to exactly fill a padding with blank spaces, digits, and a decimal point.
Paraphrasing the data and code I have now
floats = [321.1234561, 21.1234561, 1.1234561, 0.123456, 0.02345, 0.0034, 0.0004567]
for number in floats:
print('{:>8.6f}'.format(number))
This outputs
321.123456
21.123456
1.123456
0.123456
0.02345
0.0034
0.000457
I am looking for a way to print the following in a for loop assuming I don't know the amount of digits that will be to the left of the decimal place and the number of digits to the left never exceeds the padding which is 8 for this example.
321.1234
21.12345
1.123456
0.123456
0.02345
0.0034
0.000457
Similar questions have been asked about printing floating points with a certain width but the width they were talking about appeared to be the precision rather than the total number of character used to print the number.
Edit:
I have added a number to the end of the list for the following reason. The use of the specifier 'g' with 7 significant figures was recommended by attdona. This prevents the padding from being exceeded for numbers greater than or equal to 1 but not for numbers less than 1 with precision greater than 6. Using {:>8.7g} instead gives
321.1234
21.12345
1.123456
0.123456
0.02345
0.0034
0.0004567
Where the only one that exceeds the padding is the newly added one.
Use the General format type specifier g:
'{:>8.7g}'.format(number)
reference: https://docs.python.org/3/library/string.html#format-specification-mini-language
Update: For small numbers this format fails to align correctly. In this case you may adopt a mixed approach, but keep in mind that very small numbers will round to zero
for number in floats:
fstr = '{:>8.7g}'.format(number)
if len(fstr) > 8:
fstr = '{:>8.6f}'.format(number)
print(fstr)
for i in floats:
print('{:>8}'.format(f'{i:{8}.{8-len(str(int(i)))-1}f}'.rstrip('0')))
321.1235
21.12346
1.123456
0.123456
0.02345
0.0034
Starting with a list of integers the task is to convert each integer into a string such that the resulting list of strings will be in numeric order when sorted lexicographically.
This is needed so that a particular system that is only capable of sorting strings will produce an output that is in numeric order.
Example:
Given the integers
1, 23, 3
we could convert the to strings like this:
"01", "23", "03"
so that when sorted they become:
"01", "03", "23"
which is correct. A wrong result would be:
"1", "23", "3"
because that list is sorted in "string order", not in numeric order.
I'm looking for something more efficient than the simple zero-padding scheme. In order to cover all possible 32 bit integers we'd need to pad to 10 digits which is inefficient.
For integers, prefix each number with the length. To make it more readable, use 'a' for length 1, and 'b' for length 2. Example:
non-encoded encoded
1 "a1"
3 "a3"
23 "b23"
This scheme is a bit simpler than prefixing each digit, but only works with numbers, not numbers mixed with text. It can be made to work for negative numbers as well, and even BigDecimal numbers, using some tricks. I wrote an implementation in Apache Jackrabbit 2.x, to make BigDecimal indexable (sortable) as text. For that, I used a format that only uses the characters '0' to '9' and consists of:
one character for: signum(value) + 2
one character for: signum(exponent) + 2
one character for: length(exponent) - 1
multiple characters for: exponent
multiple characters for: value (-1 if inverted)
Only the signum is encoded if the value is zero. The exponent is not encoded if zero. Negative values are "inverted" character by character (0 => 9, 1 => 8, and so on). The same applies to the exponent.
Examples:
non-encoded encoded
0 "2"
2 "322" (signum 1; exponent 0; value 2)
120 "330212" (signum 1; exponent signum 1, length 1, value 2; value 12)
-1 "179" (signum -1, rest inverted; exponent 0; value 1 (-1, inverted))
Values between BigDecimal(BigInteger.ONE, Integer.MIN_VALUE) and BigDecimal(BigInteger.ONE, Integer.MAX_VALUE) are supported.
TL;DR
Encode digits according to their order of magnitude (OM) and other characters so they sort as desired, relative to numbers: jj-a123 would be encoded zjzjz-zaC1B2A3
Longer explanation
This would depend somewhat upon the sorting algorithm that will finally be used to sort and how one would want any given punctuation characters to be sorted in relation to letters and numbers, but if it's "ascii-betical" or similar, you could encode each digit of a number to represent its order of magnitude (OM) in the number, while encoding other characters such that they would sort according to your desired sort order.
For simplicity, I would suggest beginning with encoding every non-numeric character with a "high" value (e.g. lower case z or even ~ if final value is ASCII), so that it sorts after encoded digits. Then cache each digit encountered until another non-numeric is encountered, then encode each cached digit with a value representing its OM. If the number 12945 was encountered in between non-numerics, you would output an E to encode an OM of 5, then the digit that is that order of magnitude, 1, followed by the next OM of 4 (D) and its associated digit, 2. Continue until all numeric digits have been flushed, then continue with non-numerics.
Non-numerics would be treated individually and ranked relative to the OM of digits. If it is desired for them to sort "above" numbers (perhaps the space character or certain others deemed special) they would be encoded by prepending a low-value character (like the space character, if final value will be treated and sorted as ASCII). When/if another numeric is encountered, begin caching and encode according to OM once all consecutive numerics are cached.
Alternately, processing the string in reverse order would preclude the need to cache numbers except for a single "is it a digit?" test and "is the last character a digit?" test. If the first is not true, then use (one of?) the "non-digit" OM character(s). If the first test is true then use the lowest-OM "digit" character (A in my examples). If both tests are true, then increment your OM character (A -> B or E -> F) before use.
Certain levels of additional filtering - or even translation - could be applied. If one wanted to allow accurate sorting based upon Roman numerals, one could encode them as decimal (or even hexadecimal) numbers with an appropriate OM.
Treating decimal points (either periods or commas, depending) as actual decimal separators, and distinct from other punctuation would probably be beyond the true utility of this encoding scheme, as alphanumeric fields seldom use a period or comma as a decimal separator. If it is desired to use them that way, the algorithm would simply detect a decimal separator (either period or comma as appropriate, in between digits) and not encode the numeric portion after that separator as anything but normal text. Fractional portions are actually sorted correctly during a normal ASCII based sort, because more digits represents greater precision - not greater magnitude.
Examples
non-encoded encoded
----------- -------
12345 E1D2C3B4A5
a100 zaC1B0A0
a20 zaB2A0
a2000 zaD2C0B0A0
x100.5 zxC1B0A0z.A5
x100.23 zxC1B0A0z.B2A3
1, 23, 3 A1z,z B2A1z,z A3
1, 2, 3 A1z,z A2z,z A3
1,2,3 A1z,A2z,A3
Potential advantages
Going somewhat beyond simple numeric sorting, some advantages to this encoding method would be several aspects of flexibility with final effective sort order - you are essentially encoding a category for each character - digits get a category based upon their position within the greater string of digits known as a number, while other characters are simply told to sort in their normal way (e.g. ASCII), but after numbers. Any exceptions that should sort before numbers or in other orders would be in one or more additional categories. ASCII can effectively be re-encoded to sort in a non-ASCII way:
You could encode lower case letters to sort before or along with upper case letters. To switch the lower and upper cases, you encode lower case letters with a y and upper case letters with a z. For a pseudo-case-insensitive sort, categorizing both A and a with the same encoding character would sort both of them before B and b, though A would nonetheless always sort before a
If you want Extended ASCII characters (e.g. with diacritics) to sort along with their ASCII cousins, you encode À, Á, Â, Ã, Ä, Å, and Æ along with A by using an a as the OM character, encode B, C, and Ç with a b, and E, È, É, Ê, and Ë with a c, etc. The same intra-category sort order caveat still applies, and some decisions need to be made on characters like capital Eth, and to a certain extent others like Thorn, and Sharp S (Ð, Þ, and ß respectively) as to whether they will sort based on similarities in appearance or pronunciation, or instead more properly perhaps, alphabetical order.
Small advantage of being basically human-readable, with effort
Caveats
Though this allows many 'categories' of characters to be defined, be sure to remember that each order of magnitude for digits is its own category - you need to know that the data will not contain numbers that are greater in OM than approximately 250, depending upon how many other categories you wish to define (ASCII 0 is reserved for storing strings, and there needs to be at least one other character to indicate "not a digit" - at least for alphanumeric data - making the maximum perhaps 254 orders of magnitude), but that should be plenty for any situation I can imagine. I'm not sure what other issues quantum computing will bring about, but there's probably a quantum solution to it, whatever it is.
Finally, if the hyphen is encoded as a non-numeric character, and all non-numerics are encoded with a higher OM than digits, negative numbers would be encoded as greater than any positive number. The hyphen should be encoded as a lower-than-digit-OM (perhaps only when preceding a digit) if negative numbers need to be sorted correctly according to magnitude.
Since the ASCII code of A is greater than 9, you could encode them as hexadecimal strings.
The integers
1, 23, 3
can be encoded as
00000001, 00000017, 00000003
and 32-bit integers can always be encoded as 8-character strings. (assume unsigned)
I am not looking for help with my homework. I just need someone to show me the direction to do it.
I know the answer theoretically. I just stuck with idea of how to prove it mathematically.
here is the question.
Representing a number in the octal system require, on the average, about 10 percent more characters than in the decimal system.
How can I prove this mathematically?
Suppose you wanted to represent a given number x in both systems. In the decimal system, this will take in the order of log10(x) digits. In the octal system, it will take in the order of log8(x) digits.
For any a and b, loga(b) can be written as logc(b)/logc(a) for a given c. In particular, let c=10. Therefore, log8(x) = log10(x)/log10(8) ~= 1.1 log10(x), which means log8(x) is about 1.1 times greater than log10(x) for any given x. Note that this result is exact aside from the rounding. What is not exact is approximating the number of digits by log10(x) and log8(x).
The approximative number of decimal digits required for representing a number is : log10(x), and the number of octal digits is : log8(x)
Which means that the average ratio is log8(x)/log10(x)
As log8(x) = ln(x)/ln(8) and log10(x) = ln(x)/ln(10)
The average ratio is ln(10)/ln(8) = 1.1073...
Of course this is not a 100% exact demonstration, a real demonstration would define exactly the number we are trying to find (such as the average number of digits for numbers between 0 and n when n goes to infinity, etc...) and would compute the exact number of digits (which is an integer) and not an approximation.
Suppose I want to conver the number 0.011124325465476454 to string in MATLAB.
If I hit
mat2str(0.011124325465476454,100)
I get 0.011124325465476453 which differs in the last digit.
If I hit num2str(0.011124325465476454,'%5.25f')
I get 0.0111243254654764530000000
which is padded with undesirable zeros and differs in the last digit (3 should be 4).
I need a way to convert numerics with random number of decimals to their EXACT string matches (no zeros padded, no final digit modification).
Is there such as way?
EDIT: Since I din't have in mind the info about precision that Amro and nrz provided, I am adding some more additional info about the problem. The numbers I actually need to convert come from a C++ program that outputs them to a txt file and they are all of the C++ double type. [NOTE: The part that inputs the numbers from the txt file to MATLAB is not coded by me and I'm actually not allowed to modify it to keep the numbers as strings without converting them to numerics. I only have access to this code's "output" which is the numerics I'd like to convert]. So far I haven't gotten numbers with more than 17 decimals (NOTE: consequently the example provided above, with 18 decimals, is not very indicative).
Now, if the number has 15 digits eg 0.280783055069002
then num2str(0.280783055069002,'%5.17f') or mat2str(0.280783055069002,17) returns
0.28078305506900197
which is not the exact number (see last digits).
But if I hit mat2str(0.280783055069002,15) I get
0.280783055069002 which is correct!!!
Probably there a million ways to "code around" the problem (eg create a routine that does the conversion), but isn't there some way using the standard built-in MATLAB's to get desirable results when I input a number with random number of decimals (but no more than 17);
My HPF toolbox also allows you to work with an arbitrary precision of numbers in MATLAB.
In MATLAB, try this:
>> format long g
>> x = 0.280783054
x =
0.280783054
As you can see, MATLAB writes it out with the digits you have posed. But how does MATLAB really "feel" about that number? What does it store internally? See what sprintf says:
>> sprintf('%.60f',x)
ans =
0.280783053999999976380053112734458409249782562255859375000000
And this is what HPF sees, when it tries to extract that number from the double:
>> hpf(x,60)
ans =
0.280783053999999976380053112734458409249782562255859375000000
The fact is, almost all decimal numbers are NOT representable exactly in floating point arithmetic as a double. (0.5 or 0.375 are exceptions to that rule, for obvious reasons.)
However, when stored in a decimal form with 18 digits, we see that HPF did not need to store the number as a binary approximation to the decimal form.
x = hpf('0.280783054',[18 0])
x =
0.280783054
>> x.mantissa
ans =
2 8 0 7 8 3 0 5 4 0 0 0 0 0 0 0 0 0
What niels does not appreciate is that decimal numbers are not stored in decimal form as a double. For example what does 0.1 look like internally?
>> sprintf('%.60f',0.1)
ans =
0.100000000000000005551115123125782702118158340454101562500000
As you see, matlab does not store it as 0.1. In fact, matlab stores 0.1 as a binary number, here in effect...
1/16 + 1/32 + 1/256 + 1/512 + 1/4096 + 1/8192 + 1/65536 + ...
or if you prefer
2^-4 + 2^-5 + 2^-8 + 2^-9 + 2^-12 + 2^13 + 2^-16 + ...
To represent 0.1 exactly, this would take infinitely many such terms since 0.1 is a repeating number in binary. MATLAB stops at 52 bits. Just like 2/3 = 0.6666666666... as a decimal, 0.1 is stored only as an approximation as a double.
This is why your problem really is completely about precision and the binary form that a double comprises.
As a final edit after chat...
The point is that MATLAB uses a double to represent a number. So it will take in a number with up to 15 decimal digits and be able to spew them out with the proper format setting.
>> format long g
>> eps
ans =
2.22044604925031e-16
So for example...
>> x = 1.23456789012345
x =
1.23456789012345
And we see that MATLAB has gotten it right. But now add one more digit to the end.
>> x = 1.234567890123456
x =
1.23456789012346
In its full glory, look at x, as MATLAB sees it:
>> sprintf('%.60f',x)
ans =
1.234567890123456024298320699017494916915893554687500000000000
So always beware the last digit of any floating point number. MATLAB will try to round things intelligently, but 15 digits is just on the edge of where you are safe.
Is it necessary to use a tool like HPF or MP to solve such a problem? No, as long as you recognize the limitations of a double. However tools that offer arbitrary precision give you the ability to be more flexible when you need it. For example, HPF offers the use and control of guard digits down in that basement area. If you need them, they are there to save the digits you need from corruption.
You can use Multiple Precision Toolkit from MATLAB File Exchange for arbitrary precision numbers. Floating point numbers do not usually have a precise base-10 presentation.
That's because your number is beyond the precision of the double numeric type (it gives you between 15 to 17 significant decimal digits). In your case, it is rounded to the nearest representable number as soon as the literal is evaluated.
If you need more precision than what the double-precision floating-points provides, store the numbers in strings, or use arbitrary-precision libraries. For example use the Symbolic Toolbox:
sym('0.0111243254654764549999999')
You cannot get EXACT string since the number is stored in double type, or even long double type.
The number stored will be a subtle more or less than the number you gives.
computer only knows binary number 0 & 1. You must know that numbers in one radix may not expressed the same in other radix. For example, number 1/3, radix 10 yields 0.33333333...(The ellipsis (three dots) indicate that there would still be more digits to come, here is digit 3), and it will be truncated to 0.333333; radix 3 yields 0.10000000, see, no more or less, exactly the amount; radix 2 yields 0.01010101... , so it will likely truncated to 0.01010101 in computer,that's 85/256, less than 1/3 by rounding, and next time you fetch the number, it won't be the same you want.
So from the beginning, you should store the number in string instead of float type, otherwise it will lose precision.
Considering the precision problem, MATLAB provides symbolic computation to arbitrary precision.
I know that most programming languages have functions built in for doing that for you, but how do those functions work?
The javadoc about the Double toString() method is quite comprehensive:
Creates a string representation of the double argument. All characters mentioned below are ASCII characters.
If the argument is NaN, the result is the string "NaN".
Otherwise, the result is a string that represents the sign and magnitude (absolute value) of the argument. If the sign is negative, the first character of the result is '-' ('-'); if the sign is positive, no sign character appears in the result. As for the magnitude m:
If m is infinity, it is represented by the characters "Infinity"; thus, positive infinity produces the result "Infinity" and negative infinity produces the result "-Infinity".
If m is zero, it is represented by the characters "0.0"; thus, negative zero produces the result "-0.0" and positive zero produces the result "0.0".
If m is greater than or equal to 10^-3 but less than 10^7, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' (.), followed by one or more decimal digits representing the fractional part of m.
If m is less than 10^-3 or not less than 10^7, then it is represented in so-called "computerized scientific notation." Let n be the unique integer such that 10^n<=m<10^(n+1); then let a be the mathematically exact quotient of m and 10^n so that 1<=a<10. The magnitude is then represented as the integer part of a, as a single decimal digit, followed by '.' (.), followed by decimal digits representing the fractional part of a, followed by the letter 'E' (E), followed by a representation of n as a decimal integer, as produced by the method Integer.toString(int).
How many digits must be printed for the fractional part of m or a? There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double. That is, suppose that x is the exact mathematical value represented by the decimal representation produced by this method for a finite nonzero argument d. Then d must be the double value nearest to x; or if two double values are equally close to x, then d must be one of them and the least significant bit of the significand of d must be 0.
Is that enough? Otherwise you might like to look up the implementation too...
A simple (but non-generic, naïve and slow way):
convert the number to an integer, then divide this value by 10 stepwise to find out its digits in reverse order. Concatenate them together and you have the integer representation.
substract the integer from the original number, now multiply by 10 stepwise and find the digits after the decimal point. Concatenate the first string with a point and this second string.
This has a few problems, of course:
slow as hell;
doesn't work for negative numbers;
won't give you exponential notation for very small or large numbers.
All in all, it's an idea, but not a very good one; I suspect there are no programming languages that do this.
This paper by Guy Steele provides details on how to do this correctly. It's much more subtle than you might think.
http://portal.acm.org/citation.cfm?id=93559
"Printing Floating-Point Numbers Quickly and Accurately" - Robert G. Burger
Scheme and C code for above.
As Oded mentioned in a comment, different languages will do this in different ways. As an example, here's how Ruby 1.9 does it (in C). Your best bet, just as a research exercise, will be to look into open-source languages and see how they do it.