I have tried the following method and it worked.
a=[1,2,3]
b=[5,6,7]
for i=1:3
a=replace(a,a[i]=>b[i]*a[i])
end
The result showed: a=[5,12,21], which is the product I wanted,elementwise product.
However, I tried to use the same method for getting the product I want but it didn't work.
a=[]
for i=1:10
a=push!(a,2^i)
end
for i=1:10
a=replace(a,a[i]=>a[i]*a[i])
end
But the result is
a=[65536,65536,4096,65536,1048576,4096,16384,65536,262144,1048576]
And I want to yield
a=[4,16,64,256,1024,4096,16384,65536,262144,1048576]
The problem here is, that replace might not do what you want. The command
replace(A, old => new)
takes a collection A and creates a new collection where every occurrence of old is replaced by new.
So if we look at your example, in the first iteration we replace every occurrence of a[1] == 2 by 4. This yields
a == [4, 4, 8, 16, 32, 64, 128, 256, 512, 1024]
In the second iteration, we replace every occurrence of a[2] == 4 by 16. This yields
a == [16, 16, 8, 16, 32, 64, 128, 256, 512, 1024]
and so on. This should explain why you get that weird result.
Apart from the broadcasts a .= a .* a or a .= a .^ 2 that Oscar Smith mentioned in his comment, you could also use the functions map
a = map(x -> x^2, a)
or map!:
map!(x -> x^2, a, a)
The difference between map and map! is, that map creates a new version and map! writes to an already existing collection. In this example, the input collection is the same as the output collection.
Related
Draw below matrix using c++. Problem require a function, which could be called into the main().
x!x!x
~~~~~
x!x!x
~~~~~
x!x!x
I tried comparing the location 0,2,4. and tried to print but is there any other way to do this problem ?
If the matrix are characters, you could do something like this:
char board[] =
"x|x|x\n"
"-+-+-\n"
"x|x|x\n"
"-+-+-\n"
"x|x|x\n"
;
The columns containing the character 'x' are located at indices 0, 2, 4, 14, 16, 18, 26, 28, 30. Row indices are 0, 14, and 28.
Hint: there are 6 characters per row.
Hint: columns indices are (row * (characters per row)) + ((column - 1) * (2 characters per row))
This has the nice benefit of only requirement one statement to print:
std::cout.write(&board[0], sizeof(board) - 1U);
The - 1U is so that the terminating nul is not sent to cout.
I am new with Mathematica and I have one more task to figure out, but I can't find the answer. I have two lists of numbers ("b","u"):
b = {8.734059001373602`, 8.330508824111284`, 5.620669156438947`,
1.4722145583571766`, 1.797504620275392`, 7.045821078656974`,
2.1437334927375247`, 2.295629405840401`, 9.749038328921163`,
5.9928406294151095`, 5.710839663259195`, 7.6983109942364365`,
1.02781847368645`, 4.909108426318685`, 2.5860897177525572`,
9.56334726886076`, 5.661774934433563`, 3.4927397824800384`,
0.4570000499566351`, 6.240122061193738`, 8.371962670138991`,
4.593105388706549`, 7.653068139076581`, 2.2715973346475877`,
7.6234743784167875`, 0.9177107503732636`, 3.182296027902268`,
6.196168580445633`, 0.1486794884986935`, 1.2920960388213274`,
7.478757220079665`, 9.610332785387424`, 0.05088141346751485`,
3.940557901075696`, 5.21881311050797`, 7.489624788199514`,
8.773397599406234`, 3.397275198258715`, 1.4847171141876618`,
0.06574278834161795`, 0.620801320529969`, 2.075457888143216`,
5.244608900551409`, 4.54384757203616`, 7.114276285060143`,
2.8878711430358344`, 5.70657733453041`, 8.759173986432632`,
1.9392596667256967`, 7.419234634325729`, 8.258205508179927`,
1.185315253730261`, 3.907753644335596`, 7.168561412289151`,
9.919881985898002`, 3.169835543867407`, 8.352858871046699`,
7.959492335118693`, 7.772764587074317`, 7.091413185764939`,
1.433673058797801`};
and
u={5.1929, 3.95756, 5.55276, 3.97068, 5.67986, 4.57951, 4.12308,
2.52284, 6.58678, 4.32735, 7.08465, 4.65308, 3.82025, 5.01325,
1.17007, 6.43412, 4.67273, 3.7701, 4.10398, 2.90585, 3.75596,
5.12365, 4.78612, 7.20375, 3.19926, 8.10662};
This is the LinePlot of "b" and "u";
I need to compare first 5 numbers from "b" to 1st number in "u" and always leave the maximum (replace "b"<"u" with "u"). Then I need to shift by 2 numbers and compare 3rd, 4th, 5th, 6th and 7th "b" with 2nd "u" and so on (shift always => 2 steps). But the overlapping numbers need to be "remembered" and compared in the next step, so that always the maximum is picked (e.g. 3rd, 4th and 5th "b" has to be > than 1st and 2nd "u").
Possibly the easiest way would be to cover the maximums showed in the image throughout the whole function, but I am new to this software and I don't have the experience to do that. Still It would be awesome if someone would figure out how to do this with a function that would do what I have described above.
I believe this does what you want:
With[{n = Length # u},
Array[b[[#]] ~Max~ Take[u, ⌊{#-2, #+1}/2⌋ ~Clip~ {1, n}] &, 2 n + 3]
]
{8.73406, 8.33051, 5.62067, 5.1929, 5.55276, 7.04582, 5.55276, 5.55276, 9.74904,--
Or if the length of u and v are appropriately matched:
With[{n = Length # u},
MapIndexed[# ~Max~ Take[u, ⌊(#2[[1]] + {-2, 1})/2⌋ ~Clip~ {1, n}] &, b]
]
These are quite a lot faster than Mark's solution. With the following data:
u = RandomReal[{1, 1000}, 1500];
b = RandomReal[{1, 1000}, 3004];
Mark's code takes 2.8 seconds, while mine take 0.014 and 0.015 seconds.
Please ask your future questions on the dedicated Mathematica StackExchange site:
I think that there's a small problem with your data, u doesn't have as many elements as Partition[b,5,2]. Leaving that to one side, the best I could do was:
Max /# Transpose[
Table[Map[If[# > 0, Max[#, u[[i]]], 0] &,
RotateRight[PadRight[Partition[b, 5, 2][[i]], Length[b]],
2 (i - 1)]], {i, 1, Length[u]}]]
which starts producing the same numbers as in your comment.
As ever, pick this apart from the innermost expression and work outwards.
Given an array of 20 numbers, I would like to extract all possible combinations of two groups, with ten numbers in each, order is not important.
combinations([1, 2, 3], 2)
in Julia will give me all possible combinations of two numbers drawn from the array, but I also need the ones that were not drawn...
You can use setdiff to determine the items missing from any vector, e.g.,
y = setdiff(1:5, [2,4])
yields [1,3,5].
After playing around for a bit, I came up with this code, which seems to work. I'm sure it could be written much more elegantly, etc.
function removeall!(remove::Array, a::Array)
for i in remove
if in(i, a)
splice!(a, indexin([i], a)[1])
end
end
end
function combinationgroups(a::Array, count::Integer)
result = {}
for i in combinations(a, count)
all = copy(a)
removeall!(i, all)
push!(result, { i; all } )
end
result
end
combinationgroups([1,2,3,4],2)
6-element Array{Any,1}:
{[1,2],[3,4]}
{[1,3],[2,4]}
{[1,4],[2,3]}
{[2,3],[1,4]}
{[2,4],[1,3]}
{[3,4],[1,2]}
Based on #tholy's comment about instead of using the actual numbers, I could use positions (to avoid problems with numbers not being unique) and setdiff to get the "other group" (the non-selected numbers), I came up with the following. The first function grabs values out of an array based on indices (ie. arraybyindex([11,12,13,14,15], [2,4]) => [12,14]). This seems like it could be part of the standard library (I did look for it, but might have missed it).
The second function does what combinationgroups was doing above, creating all groups of a certain size, and their complements. It can be called by itself, or through the third function, which extracts groups of all possible sizes. It's possible that this could all be written much faster, and more idiomatical.
function arraybyindex(a::Array, indx::Array)
res = {}
for e in indx
push!(res, a[e])
end
res
end
function combinationsbypos(a::Array, n::Integer)
res = {}
positions = 1:length(a)
for e in combinations(positions, n)
push!(res, { arraybyindex(a, e) ; arraybyindex(a, setdiff(positions, e)) })
end
res
end
function allcombinationgroups(a::Array)
maxsplit = floor(length(a) / 2)
res = {}
for e in 1:5
println("Calculating for $e, so far $(length(res)) groups calculated")
push!(res, combinationsbypos(a, e))
end
res
end
Running this in IJulia on a 3 year old MacBook pro gives
#time c=allcombinationgroups([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20])
println(length(c))
c
Calculating for 1, so far 0 groups calculated
Calculating for 2, so far 20 groups calculated
Calculating for 3, so far 210 groups calculated
Calculating for 4, so far 1350 groups calculated
Calculating for 5, so far 6195 groups calculated
Calculating for 6, so far 21699 groups calculated
Calculating for 7, so far 60459 groups calculated
Calculating for 8, so far 137979 groups calculated
Calculating for 9, so far 263949 groups calculated
Calculating for 10, so far 431909 groups calculated
elapsed time: 11.565218719 seconds (1894698956 bytes allocated)
Out[49]:
616665
616665-element Array{Any,1}:
{{1},{2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}}
{{2},{1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}}
⋮
{{10,12,13,14,15,16,17,18,19,20},{1,2,3,4,5,6,7,8,9,11}}
{{11,12,13,14,15,16,17,18,19,20},{1,2,3,4,5,6,7,8,9,10}}
ie. 53,334 groups calculated per second.
As a contrast, using the same outer allcombinationgroups function, but replacing the call to combinationsbypos with a call to combinationgroups (see previous answer), is 10x slower.
I then rewrote the array by index group using true or false flags as suggested by #tholy (I couldn't figure out how to get it work using [], so I used setindex! explicitly, and moved it into one function. Another 10x speedup! 616,665 groups in 1 second!
Final code (so far):
function combinationsbypos(a::Array, n::Integer)
res = {}
positions = 1:length(a)
emptyflags = falses(length(a))
for e in combinations(positions, n)
flag = copy(emptyflags)
setindex!(flag, true, e)
push!(res, {a[flag] ; a[!flag]} )
end
res
end
function allcombinationgroups(a::Array)
maxsplit = floor(length(a) / 2)
res = {}
for e in 1:maxsplit
res = vcat(res, combinationsbypos(a, e))
end
res
end
This code was on an exam and it asked what it's output was going to be.
I got it wrong unfortunately and put it was all 1's.
I'm a little confused with what this program is doing specifically with the if/else statement.
I'm a C programmer, so if possible could someone please translate the if/else statement into C code so I can understand what is going on. Thank you!
EDIT: to clarify, I'm not sure what the condition means "if x in d"
def somefunction(L):
d = {}
for x in L:
if x in d:
d[x] = d[x] + 1
else:
d[x] = 1
return d
L = [6, 10, -2, 2, 6, 4, -2, 6]
print somefunction(L)
output: {10: 1, 2: 1, 4: 1, -2: 2, 6: 3}
in in Python performs a containment check. It looks at the right-hand operand to see if it contains the left-hand operand.
>>> 2 in [1, 2, 4]
True
>>> 3 in [1, 2, 4]
False
I'd encourage you NOT to translate everything into C. Python is considerably different and trying to keep things in a C frame of mind will make things harder to understand.
One thing that is great is that Python is interpreted, so you can type "python" and then enter commands to see what they do. You can exam all the variables as things are manipulated. For example, you can do:
L = [6, 10, -2, 2, 6, 4, -2, 6]
for x in L:
print x
To see what the "in" does. Likewise for the rest of the code. Also, there are great online tutorials on Python, Google "Dive into Python", for example.
See Basically in this code what you are doing is you are making a count of no of times the element is repeated in the list..you are using dictionary as a means to take the count..
First of all in the if-else block you are checking whether the element is present or not..if its present then you are incrementing the count using the element as key..else you are creating a new key,key being the element and default value being 1...
Thus you iterate all over the list and check the count of each element in the list..
d[i]=j
#i is key,j is value.
And at last you print your findings by printing the dictionary..!!
The output from Mathematica with the following operation FactorInteger[28851680048402838857] is as follows:
{{3897424303, 1}, {7402755719, 1}}
My question is: how could I go about extracting the two prime numbers (without the exponents) and assign them to an arbitrary variable?
I basically want to retrieve two primes, whatever they may be, and assign them some variables.
Ex: x0 = 3897424303 and x1 = 7402755719
Thanks!
The output is a list and you can use list manipulating functions like Part ([[ ]]) to pick the pieces you want, e.g.,
{x0, x1} = FactorInteger[28851680048402838857][[All, 1]]
or, without Part:
{{x0,dummy}, {x1,dummy}} = FactorInteger[28851680048402838857];
Implicit in your question is the issue of handing parts of the expression that is returned as output from functions such as FactorInteger. Allow me to suggest alternatives.
1. Keep all of the values in a {list} and access each element with Part:
x = First /# FactorInteger[7813426]
{2, 31, 126023}
x[[1]]
x[[3]]
2
126023
2. Store factors as values of the function x, mimicking indexation of an array:
(This code uses MapIndexed, Function.)
Clear[x]
MapIndexed[
(x[First##2] = First##1) &,
FactorInteger[7813426]
];
x[1]
x[3]
2
126023
You can see all the values using ? or ?? (see Information):
?x
Global`x
x[1]=2
x[2]=31
x[3]=126023