I have a very long expression that I need to take the derivative of:
D(expression(-4750000+(((14400*(((x/25)*1)-7.2))+(0*((x*1.05)-7.2))+(144*
((x*0.6)-7.2))*30.41667)-2500)/((0.1+1)^((1-0.5)/12))+(((13216.5802942644*
(((x/25)*1)-7.2))+(0*((x*1.05)-7.2))+(132.165802942644*
((x*0.6)-7.2))*30.41667)-2500)/((0.1+1)^((2-0.5)/12))+.........),'x')
When the expression is less than 4000 characters I get a solution. When it is more than 4000 characters, R gives me a newline and +, expecting more input. I haven't been able to find documentation on a character limit. Does anyone know why this might be? Any workarounds or alternatives to finding this derivative? The final character length will be at least 50k.
You are probably seeing an effect of input through stdin which uses the OS newline function. Try putting it in a text file and source()-ing.
Related
I have spend hours to look for a proper solutions but I found nothing on Internet. There is my question. In R, I have a specific list of characters containings my desired variable names ("2011_Q4", "2012_Q1", ...). When I try to assign a dataset to each of this name with a loop, it does work but the output it's strange. Indeed, I have
> View(`2011_Q4`)
instead of
> View(2011_Q4)
And I don't know how to remove this apostrophe. It's very annoying since I have to type this ` in order to call the variable.
Somebody can help me? I would appreciate his help.
Thanks a lot and best regards
Firstly, it's a backtick (`), not an apostrophe ('). In R, backticks occasionally denote variable names; apostrophes work as single quotes for denoting strings.
The issue you're having is that your variables start with a number, which is not allowed in R. Since you somehow made it happen anyway, you need to use backticks to tell R not to interpret 2011_Q4 as a number, but as a variable.
From ?Quotes:
Names and Identifiers
Identifiers consist of a sequence of letters, digits, the period (.)
and the underscore. They must not start with a digit nor underscore,
nor with a period followed by a digit. Reserved words are not valid
identifiers.
The definition of a letter depends on the current locale, but only
ASCII digits are considered to be digits.
Such identifiers are also known as syntactic names and may be used
directly in R code. Almost always, other names can be used provided
they are quoted. The preferred quote is the backtick (`), and deparse
will normally use it, but under many circumstances single or double
quotes can be used (as a character constant will often be converted to
a name). One place where backticks may be essential is to delimit
variable names in formulae: see formula.
The best solution to your issue is simply to change your variable names to something that starts with a character, e.g. Y2011_Q4.
I have a project about Visible Light Communication. I need to get user input (e.g sentence) and convert this sentence into binary values so that I can turn on-off the LED.
So far I managed to transmit data but i am having issue with how to convert this string into binary.
One of the ideas that i came up with is to convert this string into 1D array with 1s and 0s. Each 8 binary element in the array represents the ascii equvialant. However i couldnt manage to achieve this task.
if you guys give me another method to complete this, it'll be greatly appriciated.
I'm sure there is a better way of doing this, but i couldnt find it so far.
If the text is only A-Z and a-z i.e. Without any special characters like spaces, periods, commas, $&()#, etc. then you can encode them with your own encoding e.g. A=0 B=1 .... a= 26 ... z = 51. Doing this, you will need 6 bits to represent each character.
On the other hand, you could use more advanced algorithms like Huffman coding to encode each character.
However, if you're having problems with transmitting the ASCII representation, you'll have a problem with those two options. I'd say stick with the ASCII encoding till you have a stronger reason to try something else.
I am reading a csv file with some really big numbers like 1327707999760, but R automatically converts it into 1.32771e+12. I've tried to assign it a double class but it didn't work because it's already a rounded value.
I've checked other posts like Preserving large numbers . People said "It's not in a "1.67E+12 format", it just won't print entirely using the defaults. R is reading it in just fine and the whole number is there." But when I tried to do some arithmetic things on them, it's just not right.
For example:
test[1,8]
[1] 1.32681e+12
test[2,8]
[1] 1.32681e+12
test[2,8]-test[1,8]
[1] 0
But I know they are different numbers!
That's not large. It is merely a representation problem. Try this:
options(digits=22)
options('digits') defaults to 7, which is why you are seeing what you do. All twelve digits are being read and stored, but not printed by default.
Excel allows custom formats: Format/Cells/Custom and enter #0
I made the following regex:
(\d{5}|\d-\d{4}|\d{2}-\d{3}|\d{3}-\d{2}|\d{4}-\d)
And it seems to work. That is, it will match a 5 digit number or a 5 digit number with only 1 hyphen in it, but the hyphen can not be the lead or the end.
I would like a similar regex, but for a 25 digit number. If I use the same tactic as above, the regex will be very long.
Can anyone suggest a simpler regex?
Additional Notes:
I'm putting this regex into an XML file which is to be consumed by an ASP.NET application. I don't have access to the .net backend code. But I suspect they would do something liek this:
Match match = Regex.Match("Something goes here", "my regex", RegexOptions.None);
You need to use a lookahead:
^(?:\d{25}|(?=\d+-\d+$)[\d\-]{26})$
Explanation:
Either it's \d{25} from start to end, 25 digits.
Or: it is 26 characters of [\d\-] (digits or hyphen) AND it matched \d+-\d+ - meaning it has exactly one hyphen in the middle.
Working example with test cases
You could use this regex:
^[0-9](?:(?=[0-9]*-[0-9]*$)[0-9-]{24}|[0-9]{23})[0-9]$
The lookahead makes sure there's only 1 dash and the character class makes sure there are 23 numbers between the first and the last. Might be made shorter though I think.
EDIT: The a 'bit' shorter xP
^(?:[0-9]{25}|(?=[^-]+-[^-]+$)[0-9-]{26})$
A bit similar to Kobi's though, I admit.
If you aren't fussy about the length at all (i.e. you only want a string of digits with an optional hyphen) you could use:
([\d]+-[\d]+){1}|\d
(You may want to add line/word boundaries to this, depending on your circumstances)
If you need to have a specific length of match, this pattern doesn't really work. Kobi's answer is probably a better fit for you.
I think the fastest way is to do a simple match then add up the length of the capture buffers, why attempt math in a regex, makes no sence.
^(\d+)-?(\d+)$
This will match 25 digits and exactly one hyphen in the middle:
^(?=(-*\d){25})\d.{24}\d$
Should be validating 1-280 input characters, but it hangs when more than 280 characters are input.
Clarification
I am using the above regex to validate the length of input string to be 280 characters maximum.
I am using asp:RegularExpressionValidator to do that.
There's nothing “wrong” with it per se, but it's horrendous because with most RE engines (you don't say which one you're using) when it doesn't match with the first thing it tries because it causes the engine to backtrack and try loads of different possibilities (none of which can ever cause a match). So it's not a hang, but rather just a machine that's trying to execute around 2280 operations to see if there's a match possible. Excuse me if I don't wait around for that!
Of course, it's theoretically possible for the RE compiler to merge the (.|\s) part of the RE into something it doesn't need to backtrack to deal with. Some RE engines do this (typically the more automata-theoretic ones) but many don't (the stack-based ones).
It is trying every possible combination of . and \s for each character trying to find a version of the pattern that matches the string.
. already matches any character, so (.|\s) is redundant. Further, if you just want to check what the length of the string is, then just do that - why are you pulling out regexes?
If you really want to use a regular expression, you could use .{1, 280}$ combined with the SingleLine option, so that the . metacharacter will match everything, including new lines (see here, Regular Expression API section).