When I add a line to the middle of a file, all following lines have their number incremented.
Is there a utility that generates the list of equivalent line numbers between two files?
The output would be something like:
1 1
2 2
3 4 (line added)
4 5
One can probably create such utility by using dynamic programming in a way similar to the diff algorithm. Seems useful, hasn't already been done?
I found out it is pretty easy to do with python's difflib:
import difflib
def seq_equivs(s1, s2):
equiv = []
s = difflib.SequenceMatcher(a=s1, b=s2)
for m in s.get_matching_blocks():
if m[2] == 0:
break
for n in range(1, 1+m[2]):
equiv.append((m[0]+n, m[1]+n))
return equiv
Example usage:
f1 = open('file1.txt').read().split('\n')
f2 = open('file2.txt').read().split('\n')
for equivs in seq_equivs(f1, f2):
print('%d %d' % equivs)
Related
I am trying to take the derivative of a function including a boolean variable with sympy.
My expected result:
Two different derivatives, depending on the boolean being either True or False (i.e. 1 or 0).
Example:
import sympy as sy
c, x = sy.symbols("c x", positive=True, real=True)
bo = sy.Function("bo")
fct1 = sy.Function("fct1")
fct2 = sy.Function("fct2")
FOC2 = sy.Function("FOC2")
y = 5
a = 2
b = 4
def fct1(x):
return -0.004*x**2 + 0.25*x + 4
# the following gives the smaller positive intercept with the x-axis)
# this intercept is the threshold value for the boolean function, bo
min(sy.solve(fct1(x)-y, x))
def bo(x):
if fct1(x) <= y:
return 1
else:
return 0
def fct2(c, x):
return a + b*c + bo(x)*c
def FOC2(c, x):
return sy.diff(fct2(c, x), c)
print(FOC2(c, x))
The min-function after the comments shows me the threshold of x for bo being True or False would be 4.29..., thus positive and real.
Output:
TypeError: cannot determine truth value of Relation
I understand that the truth value depends on x, which is a symbol. Thus, without knowing x one cannot determine bo.
But how would I get my expected result, where bo is symbolic?
First off, I would advise you to carefully consider what is going on in your code the way it is pasted above. You first define a few sympy functions, e.g.
fct1 = sy.Function("fct1")
So after this, fct1 is an undefined sympy.Function - undefined in the sense that it is neither specified what its arguments are, nor what the function looks like.
However, then you define same-named functions explicitly, as in
def fct1(x):
return -0.004*x**2 + 0.25*x + 4
Note however, that at this point, fct1 ceases to be a sympy.Function, or any sympy object for that matter: you overwrite the old definition, and it is now just a regular python function!
This is also the reason that you get the error: when you call bo(x), python tries to evaluate
-0.004*x**2 + 0.25*x + 4 <= 5
and return a value according to your definition of bo(). But python does not know whether the above is true (or how to make that comparison), so it complains.
I would suggest 2 changes:
Instead of python functions, as in the code, you could simply use sympy expressions, e.g.
fct1 = -0.004*x**2 + 0.25*x + 4
To get the truth value of your condition, I would suggest to use the Heaviside function (wiki), which evaluates to 0 for a negative argument, and to 1 for positive. Its implementation in sympy is sympy.Heaviside.
Your code could then look as follows:
import sympy as sy
c, x = sy.symbols("c x", positive=True, real=True)
y = 5
a = 2
b = 4
fct1 = -0.004*x**2 + 0.25*x + 4
bo = sy.Heaviside(y - fct1)
fct2 = a + b*c + bo * c
FOC2 = sy.diff(fct2, c)
print(FOC2)
Two comments on the line
bo = sy.Heaviside(y - fct1)
(1) The current implementation does not evaluate sympy.Heaviside(0)by default; this is beacause there's differing definitions around (some define it to be 1, others 1/2). You'd want it to be 1, to be in accordance with the (weak) inequality in the OP. In sympy 1.1, this can be achieved by passing an additional argument to Heaviside, namely whatever you want Heaviside(0) to evaluate to:
bo = sy.Heaviside(y - fct1, 1)
This is not supported in older versions of sympy.
(2) You will get your FOC2, again involving a Heaviside term. What I like about this, is that you could keep working with this expression, say if you wanted to take a second derivative and so on. If, for the sake of readability, you would prefer a piecewise expression - no problem. Just replace the according line with
bo = sy.Heaviside(y - fct1)._eval_rewrite_as_Piecewise(y-fct1)
Which will translate to a piecewise function automatically. (note that under older versions, this automatically implicitly uses Heaviside(0) = 0.5 - best to use (1) and (2) together:
bo = sy.Heaviside(y - fct1, 1)._eval_rewrite_as_Piecewise(y-fct1)
Unfortunately, I don't have a working sympy 1.1 at my hands right now and can only test the old code.
One more noteconcerning sympy's piecewise functions: they are much more readable if using sympy's latex printing, by inserting
sy.init_printing()
early in the code.
(Disclaimer: I am by no means an expert in sympy, and there might be other, preferable solutions out there. Just trying to make a suggestion!)
Try to learn minizinc but after going through examples, may I just confirm that I actually have to write some procedural language if I want to get multiple output or there is a more "natural to minizinc" way to get it.
For example, suppose I want to have all distinct digits add up to 3 the answers should be 0+3 1+2 2+1 3+0 ...
My mininzinc here:
% how to generate more than one result meeting the constraints
int: n=3;
var 0..9: a;
var 0..9: b;
include "alldifferent.mzn";
constraint all_different([a, b]);
constraint a + b = n;
solve satisfy;
output [
"a + b = n \t\n",
show(a), " + ",
show(b), " = ",
show(n)];
produce only 3+0. How to get to the other answers? Thanks for any advice in advance.
I looked at a post for minizinc 1.6 and it seemed to say left out the output statement would produce all the output (Easy way to print full solution (all decision variables) in minizinc). It does not work. Only one is output.
First of all, the default is to print all variables and their values for a solution, not all solutions.
Use the option -a to get all solutions. mzn-gecode --help to see all options. In your case mzn-gecode -a test.mzn which gives:
a + b = n
3 + 0 = 3
----------
a + b = n
0 + 3 = 3
----------
a + b = n
2 + 1 = 3
----------
a + b = n
1 + 2 = 3
----------
==========
Under configuration there is an option to change the default from printing the first solution after satisfaction. Change it to user-defined-behaviour: print all solutions ... You can have output statement, btw, as well.
I am going to start illustration using a code:
A = 'G1(General G1Airlines american G1Fungus )';
Using regexp (or any other function) in Matlab I want to distinctively locate: G1, G1A and G1F.
Currently if I try to do something as:
B = regexp( A, 'G1')
It is not able to distinguish G1 with the G1A and G1F i.e. I need to force the comparison to find me only case with G1 and ignore G1A and G1F.
However, when I am searching for G1A then it should still find me the location of G1A.
Can someone please help ?
Edit: Another case for A is:
A = 'R1George Service SmalR1Al C&I)';
And the expression this time I need to find is R1 and R1A instead.
Edit:
I have a giant array containing A's and another big vector containing G1, R1, etc I need to search for.
If you want to find 'G1' but not 'G1A' or 'G1F' you can use
>> B = regexp(A, 'G1[^AF]')
B =
1
This will find 'G1' and the ^ is used to specify that it should not match any characters contained with []. Then you could use
>> B = regexp(A, 'G1[AF]')
B =
12 32
to find both 'G1A' and 'G1F'.
The given task is to call a function from within another function, where both functions are handling matrices.
Now lets call this function 1 which is in its own file:
A = (1/dot(v,v))*(Ps'*Ps);
Function 1 is called with the command:
bpt = matok(P);
Now in another file in the same folder where function 1 is located (matok.m) we make another file containing function 2 that calls function 1:
bpt = matok(P);
What I wish B to do technically, is to return the result of the following (where D is a diagonal matrix):
IGNORE THIS LINE: B = (1/dot(v,v))*(Ps'*inv(D)*Ps*inv(D);
EDIT: this is the correct B = (1/dot(v,v))*(Ps*inv(D))'*Ps*inv(D);
But B should not "re-code" what has allready been written in function 1, the challenge/task is to call function 1 within function 2, and within function 2 we use the output of function 1 to end up with the result that B gives us. Also cause in the matrix world, AB is not equal to BA, then I can't simply multiply with inv(D) twice in the end. Now since Im not allowed to write B as is shown above, I was thinking of replacing (without altering function 1, doing the manipulation within function 2):
(Ps'*Ps)
with
(Ps'*inv(D)*Ps*inv(D)
which in some way I imagine should be possible, but since Im new to Matlab have no idea how to do or where even to start. Any ideas on how to achieve the desired result?
A small detail I missed:
The transpose shouldn't be of Ps in this:
B = (1/dot(v,v))*(Ps'*inv(D))*Ps*inv(D);
But rather the transpose of Ps and inv(D):
B = (1/dot(v,v))*(Ps*inv(D))'*Ps*inv(D);
I found this solution, but it might not be as compressed as it could've been and it seems a bit unelegant in my eyes, maybe there is an even shorter way?:
C = pinv(Ps') * A
E = (Ps*inv(D))' * C
Since (A*B)' = B'*A', you probably just need to call
matok(inv(D) * Ps)
I've been given the task of translating a piece of MATLAB code into IDL and have
hit a roadblock when I came across the MATLAB function accumarry(). The
function, described here
is used to sum elements in one array, based on indices given in another. Example
1 perhaps explains this better than the actual function description at the top
of the page. In trying to reproduce Example 1 in IDL, I haven't been able to avoid a for loop, but I'm confident that it is possible. My best attempt is the following:
vals = [101,102,103,104,105]
subs = [0,1,3,1,3]
n = max(subs)+1
accum = make_array(n)
for i = 0, n-1 do begin
wVals = where(subs eq i,count)
accum[i] = count eq 0 ? 0 : total(vals[wVals])
endfor
print,accum
; 101.000 206.000 0.00000 208.000
Any advice on improving this would be greatly appreciated! I expected IDL to have a similar built-in function, but haven't been able to track one down. Perhaps some magic with histogram binning?
I found a number of possible solutions to this problem on Coyote's IDL site (not surprisingly.)
http://www.idlcoyote.com/code_tips/drizzling.html
I ended up using the following, as a compromise between performance and readability:
function accumarray,data,subs
mx = max(subs)
accum = fltarr(mx+1)
h = histogram(subs,reverse_indices=ri,OMIN=om)
for j=0L,n_elements(h)-1 do if ri[j+1] gt ri[j] then $
accum[j+om] = total(vals[ri[ri[j]:ri[j+1]-1]])
return,accum