Very new to prolog, Anyways
I have a database with names,road number,address.
habite('Name', Num, 'Address').
I want to create a rule voisins(X,Y) which returns the names of people with the same address as Y but with Num being either Num or it's adjacents(Num+1) and (Num-1).
I tried doing the following
voisins(X,Y) :-
habite(X,Num,Address),
ANum is Num+1,
habite(Y,ANum,Address);
BNum is Num-1,
habite(Y,BNum,Address).
and also tried
voisins(X,Y) :-
habite(X,Num,Address),
ANum is Num+1,
BNum is Num-1,
habite(Y,ANum,Address),
habite(Y,ANum,Address).
They both don't work. I feel like this has a very simple answer but I'm very new to Prolog so I don't really know it . Appreciate any help.
You can try the following:
voisins(X,Y) :-
habite(Y,YNum,Address),
habite(X,XNum,Address),
(XNum =:= YNum ; XNum =:= YNum - 1 ; XNum =:= YNum + 1).
The first goal finds the road number and address of person with name Y, which are then unified to the variables YNum and Address, respectively. The second goal then unifies the variable X with a name of a person that has the same address as Y, while additionally unifying XNum with this person's road number. Lastly, XNum is compared to YNum, in order to see if it satisfies the constraints you've mentioned at the beginning. This is expressed through disjunction: (;)/2.
Basically, Prolog will try to see if XNum has the same (arithmetic) value as YNum. If this check succeeds, then the predicate voisins(X,Y) is true for the name that got unified to X, so it answers it back to you and continues searching for other answers by doing backtracking. On the other hand, if it fails, it will move on to the next one and check whether that one holds, etc.
Equivalently, you could also write the above predicate as:
voisins(X,Y) :-
habite(Y,YNum,Address),
habite(X,YNum,Address).
voisins(X,Y) :-
habite(Y,YNum,Address),
habite(X,XNum,Address),
XNum =:= YNum - 1.
voisins(X,Y) :-
habite(Y,YNum,Address),
habite(X,XNum,Address),
XNum =:= YNum + 1.
Here, the disjunction has been "unfolded" into three separate definitions of voisins, where each does one of the three comparisons. The way it works is similar to the previous solution, only now Prolog will choose the next available predicate as an alternative, in case that a comparison fails. Also, notice how the XNum =:= YNum check is expressed more compactly, since the value YNum is used in habite(X,YNum,Address)to further constrain the possible values for X.
Related
Ultimately, I want to change scores of 0 to 1, scores of 1 to 2, and scores of 2 to 3. I thought one way to do that was using +1, but I realize I could also use a more complicated if then series.
Here is what I did so far:
I used the existing variable (x) to create a new variable (y=x+1) using SPSS syntax. I only want to do this for variables with values >=0 (this was my approach to excluding cells with missing data; the range for x is 0-2).
I can create x+1, but it overwrites the existing variables.
DO REPEAT x =var_1 TO var_86.
if (x>=0) x=(x+1).
end repeat.
exe.
I tried this modification, but it doesn't work:
DO REPEAT x = var_1 TO var_86 / y = var_1a TO var_86a.
IF (x >= 0) y=x +1.
END REPEAT.
EXE.
The error message is:
DO REPEAT The form VARX TO VARY to refer to a range of variables has
been used incorrectly. When using VARX TO VARY to create new
variables, X must be an integer less than or equal to the integer Y.
(Can't use A3 TO A1.)
I tried many other configurations including vectors and loops but haven't yet figured out how to do this computation across the range of variables without overwriting the existing ones. Thanks in advance for any recommendations.
The message you are getting is because SPSS doesn't understand the form var_1a TO var_86a.
For the x to y form to work the number has to be at the end of the name, so for example varA_1 to varA_86 should work.
While you're at it, here's a simple way to go about your task:
recode var_1 TO var_86 (0=1)(1=2)(2=3) into varA_1 TO varA_86.
I have been trying out some regular expressions lately. Now, I have 3 symbols a, b and c.
I first looked at a case where I don't want 2 consecutive a's. The regex would be something like:
((b|c + a(b|c))*(a + epsilon)
Now I'm wondering if there's a way to generalize this problem to say something like:
A regular expression with no two consecutive a's and no two consecutive b's. I tried stuff like:
(a(b|c) + b(a|c) + c)* (a + b + epsilon)
But this accepts inputs such as"abba" or "baab" which will have 2 consecutive a's (or b's) which is not what I want. Can anyone suggest me a way out?
If you can't do a negative match then perhaps you can use negative lookahead to exclude strings matching aa and bb? Something like the following (see Regex 101 for more information):
(?!.*(aa|bb).*)^.*$
I (think I) solved this by hand-drawing a finite state machine, then, generating a regex using FSM2Regex. The state machine is written below (with the syntax from the site):
#states
s0
s1
s2
s3
#initial
s0
#accepting
s1
s2
s3
#alphabet
a
b
c
#transitions
s0:a>s1
s0:b>s2
s0:c>s3
s1:b>s2
s1:c>s3
s2:a>s1
s2:c>s3
s3:c>s3
s3:a>s1
s3:b>s2
If you look at the transitions, you'll notice it's fairly straightforward- I have states that correspond to a "sink" for each letter of the alphabet, and I only allow transitions out of that state for other letters (not the "sink" letter). For example, s1 is the "sink" for a. From all other states, you can get to s1 with an a. Once you're in s1, though, you can only get out of it with a b or a c, which have their own "sinks" s2 and s3 respectively. Because we can repeat c, s3 has a transition to itself on the character c. Paste the block text into the site, and it'll draw all this out for you, and generate the regex.
The regex it generated for me is:
c+cc*(c+$+b+a)+(b+cc*b)(cc*b)*(c+cc*(c+$+b+a)+$+a)+(a+cc*a+(b+cc*b)(cc*b)*(a+cc*a))(cc*a+(b+cc*b)(cc*b)*(a+cc*a))*(c+cc*(c+$+b+a)+(b+cc*b)(cc*b)*(c+cc*(c+$+b+a)+$+a)+b+$)+b+a
Which, I'm pretty sure, is not optimal :)
EDIT: The generated regex uses + as the choice operator (usually known to us coders as |), which means it's probably not suitable to pasting into code. However, I'm too scared to change it and risk ruining my regex :)
You can use back references to match the prev char
string input = "acbbaacbba";
string pattern = #"([ab])\1";
var matchList = Regex.Matches(input, pattern);
This pattern will match: bb, aa and bb. If you don't have any match in your input pattern, it means that it does not contain a repeated a or b.
Explanation:
([ab]): define a group, you can extend your symbols here
\1: back referencing the group, so for example, when 'a' is matched, \1 would be 'a'
check this page: http://www.regular-expressions.info/backref.html
In a program coded in F90, I have a set of 11 vectors, each of a size (7), with names going from "S1" to "S11".
I need to be able to read a number of elements from one vector, by giving the name of this latter.
Although this problem seems an elementary one, with my beginner level, I am unable to find a way to code it...
any help ?
When a Fortran program executes it doesn't really have the information available to identify a variable based on the value of a string containing the name of a variable. You could write a sequence of if statements such as
if (mystr=='S1') x = s1
if (mystr=='S2') x = s2
...
You could pretty this up a bit with a select case construction
select case (mystr)
case ('S1')
x = s1
case ('S2')
x = s2
...
but that may not appeal much more.
A better approach by far would be to declare your vectors as elements of a rank-2 array:
real, dimension(11,7) :: s
and you can then do all sorts of computations, at run-time, to select the vector you want
myrow = an_expression_returning_an_integer_between_1_and_11_inclusive
x = s(myrow,:)
use "locationdata.dta", clear
gen ring=.
* Philly City Hall
gen lat_center = 39.9525468
gen lon_center = -75.1638855
destring(INTPTLAT10), replace
destring(INTPTLON10), replace
vincenty INTPTLAT10 INTPTLON10 lat_center lon_center , hav(distance_km) inkm
quietly su distance_km
local min = r(min)
replace ring=0 if (`min' <= distance_km < 1)
local max = ceil(r(max))
* forval loop does not work
forval i=1/`max'{
local j = `i'+1
replace ring=`i' if (`i' <= distance_km < `j')
}
I am drawing rings by 1 km from a point. The last part of the code (forval) does not work. Something wrong here?
EDIT:
The result for the forval part is as follows:
. forval i=1/`max'{
2. local j = `i'+1
3. replace ring=`i' if `i' <= distance_km < `j'
4. }
(1746 real changes made)
(0 real changes made)
(0 real changes made)
(0 real changes made)
....
So, replacing does not work for i = 2 and beyond.
A double inequality such as
(`min' <= distance_km < 1)
which makes sense according to mathematical conventions is clearly legal in Stata. Otherwise, your code would have triggered a syntax error. However, Stata does not hold evaluation until the entire expression is evaluated. The parentheses here are immaterial, as what is key is how their contents are evaluated. As it turns out, the result is unlikely to be what you want.
In more detail: Stata interprets this expression from left to right as follows. The first equality
`min' <= distance_km
is true or false and thus evaluated as 0 or 1. Evidently you want to select values such that
distance_km >= `min'
and for such values the inequality above is true and returns 1. Stata would then take a result of 1 forward and turn to the second inequality, evaluating
1 < 1
(i.e. result of first inequality < 1), but that is false for such values. Conversely,
(`min' <= distance_km < 1)
will be evaluated as
0 < 1
-- which is true (returns 1) -- if and only if
`min' > distance_km
In short, what you intend would need to be expressed differently, namely by
(`min' <= distance_km) & (distance_km < 1)
I conjecture that this is at the root of your problem.
Note that Stata has an inrange() function but it is not quite what you want here.
But, all that said, from looking at your code the inequalities and your loop both appear quite unnecessary. You want your rings to be 1 km intervals, so that
gen ring = floor(distance_km)
can replace the entire block of code after your call to vincenty, as floor() rounds down with integer result. You appear to know of its twin ceil().
Some other small points, incidental but worth noting:
You can destring several variables at once.
Putting constants in variables with generate is inefficient. Use scalars for this purpose. (However, if vincenty requires variables as input, that would override this point, but it points up that vincenty is too strict.)
summarize, meanonly is better for calculating just the minimum and maximum. The option name is admittedly misleading here. See http://www.stata-journal.com/sjpdf.html?articlenum=st0135 for discussion.
As a matter of general Stata practice, your post should explain where the user-written vincenty comes from, although it seems that is quite irrelevant in this instance.
For completeness, here is a rewrite, although you need to test it against your data.
use "locationdata.dta", clear
* Philly City Hall
scalar lat_center = 39.9525468
scalar lon_center = -75.1638855
destring INTPTLAT10 INTPTLON10, replace
vincenty INTPTLAT10 INTPTLON10 lat_center lon_center , hav(distance_km) inkm
gen ring = floor(distance_km)
Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?
The way I intend to use it is to check whether an element index in one matrix is equal to the values stored in another array (where the stored values are the indices of the elements which meet a certain criteria).
So, if the indices of the elements which meet the criteria are stored in the matrix below:
criteriacheck = [3 5 6 8 20];
Going through the main array (called array) and checking if the index matches:
for i = 1:numel(array)
if i == 'Any value stored in criteriacheck'
%# "Do this"
end
end
Does anyone have an idea of how I might go about this?
The excellent answer previously given by #woodchips applies here as well:
Many ways to do this. ismember is the first that comes to mind, since it is a set membership action you wish to take. Thus
X = primes(20);
ismember([15 17],X)
ans =
0 1
Since 15 is not prime, but 17 is, ismember has done its job well here.
Of course, find (or any) will also work. But these are not vectorized in the sense that ismember was. We can test to see if 15 is in the set represented by X, but to test both of those numbers will take a loop, or successive tests.
~isempty(find(X == 15))
~isempty(find(X == 17))
or,
any(X == 15)
any(X == 17)
Finally, I would point out that tests for exact values are dangerous if the numbers may be true floats. Tests against integer values as I have shown are easy. But tests against floating point numbers should usually employ a tolerance.
tol = 10*eps;
any(abs(X - 3.1415926535897932384) <= tol)
you could use the find command
if (~isempty(find(criteriacheck == i)))
% do something
end
Note: Although this answer doesn't address the question in the title, it does address a more fundamental issue with how you are designing your for loop (the solution of which negates having to do what you are asking in the title). ;)
Based on the for loop you've written, your array criteriacheck appears to be a set of indices into array, and for each of these indexed elements you want to do some computation. If this is so, here's an alternative way for you to design your for loop:
for i = criteriacheck
%# Do something with array(i)
end
This will loop over all the values in criteriacheck, setting i to each subsequent value (i.e. 3, 5, 6, 8, and 20 in your example). This is more compact and efficient than looping over each element of array and checking if the index is in criteriacheck.
NOTE: As Jonas points out, you want to make sure criteriacheck is a row vector for the for loop to function properly. You can form any matrix into a row vector by following it with the (:)' syntax, which reshapes it into a column vector and then transposes it into a row vector:
for i = criteriacheck(:)'
...
The original question "Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?" can be solved without any loop.
Just use the setdiff function.
I think the INTERSECT function is what you are looking for.
C = intersect(A,B) returns the values common to both A and B. The
values of C are in sorted order.
http://www.mathworks.de/de/help/matlab/ref/intersect.html
The question if i == 'Any value stored in criteriacheck can also be answered this way if you consider i a trivial matrix. However, you are proably better off with any(i==criteriacheck)