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Any Ideas on how I can accomplish this? I am very new to mathematica...
My initial thoughts were to import the data from excel in .CSV and determine the max y value in all of the sets of data, and shift the rest of the initial y values to that value. Also I need to keep the time values unchanged, and it needs to work for N# of lists.
I have no Idea how to do this, or if there is a simpler solution. Thanks!
Example:
List#1-> {{8,0},{6,1},{4,2}}, List#2-> {{7,0},{6,1},{2,2}}
List#1-> {{8,0},{6,1},{4,2}}, List#2-> {{8,0},{7,1},{3,2}}
All y values in list #2 were shifted up by one.
He has asked this question at least one other place. He seems to want to add an offset to each pair in a list so that the first of the first of all the lists become the same, the first of each pair in the rest of each list have the same offset added and the second of each pair is unchanged and he wants this to work on a relatively arbitrary number of lists.
Try to adapt this and see if it works for you
list1 = {{8, 0}, {6, 1}, {4, 2}};
list2 = {{7, 0}, {6, 1}, {2, 2}};
list3 = {{5, 0}, {7, 1}, {5, 2}};
list4 = {{9, 0}, {4, 1}, {1, 2}};
f[v_] := Map[# + {list1[[1, 1]] - v[[1, 1]], 0} &, v];
{list2, list3, list4} = Map[f, {list2, list3, list4}]
which should assign {{8,0},{7,1},{3,2}} to list2, {{8,0},{10,1},{8,2}} to list3 and {{8,0},{3,1},{0,2}} to list4.
I have read x-data (from text files) into list1, and y-data similarly into list2:
list1 = { 0.0, 0.172, 0.266, ..}
list2 = {-5.605, -5.970, -6.505, ..}
How do I combine the two lists in order to plot points {0.0, -5.605}, {0.172, -5.970}, {0.266, -6.505},....
If you don't like Pinguin Dirk's suggestion try
Transpose[{list1,list2}]
yet another..
MapThread[ {#1, #2} & , {list1, list2}]
or if you want to gracefully handle unequal length lists:
MapThread[ {#1, #2} &, Take[#, All, Min ## Length /# #] &#{list1, list2} ]
Here is another answer that creates a reusable function to pair up two vectors. The function uses a pure function that maps over the length of the shortest vector to create the pairs.
list1 = {1, 2, 3, 4, 5};
list2 = {13, 18, 20, 18, 13};
pairUp[xValues_ , yValues_] := ({xValues[[#]], yValues[[#]]}) & /#
Range[Min[Length[xValues], Length[yValues]]];
pairUp[list1, list2]
(*
{{1, 13}, {2, 18}, {3, 20}, {4, 18}, {5, 13}}
*)
Hope this helps,
Edmund
PS: New to Mathematica and hoping to improve my understanding by trying to answer a few questions on here from time to time.
Here you go
Partition[Riffle[x,y],2]
This is a basic question on list manipulation in Mathematica.
I have a large list where each element has the following schematic form: {List1, List2,Number}. For e.g.,
a = {{{1,2,3},{1,3,2},5},{{1,4,5},{1,0,2},10},{{4,5,3},{8,3,4},15}}}.
I want to make a new lists which only has some parts from each sublist. Eg., pick out the third element from each sublist to give {5,10,15} from the above. Or drop the third element to return {{{1,2,3},{1,3,2}},{{1,4,5},{1,0,2}},{{4,5,3},{8,3,4}}}.
I can do this by using the table command to construct new lists, e.g.,
Table[a[[i]][[3]],{i,1,Length[a]}
but I was wondering if there was a must faster way which would work on large lists.
In Mathematica version 5 and higher, you can use the keyword All in multiple ways to specify a list traversal.
For instance, instead of your Table, you can write
a[[All,3]]
Here Mathematica converts All into all acceptable indices for the first dimension then takes the 3rd one of the next dimension.
It is usually more efficient to do this than to make a loop with the Mathematica programming language. It is really fine for homogenous lists where the things you want to pick or scan through always exist.
Another efficient notation and shortcut is the ;; syntax:
a[[ All, 1 ;; 2]]
will scan the first level of a and take everything from the 1st to the 2st element of each sublist, exactly like your second case.
In fact All and ;; can be combined to any number of levels. ;; can even be used in a way similar to any iterator in Mathematica:
a[[ start;;end;;step ]]
will do the same things as
Table[ a[[i]], {i,start,end,step}]
and you can omit one of start, end or step, it is filled with its default of 1, Length[(of the implicit list)], and 1.
Another thing you might want to lookup in Mathematica's Help are ReplacePart and MapAt that allow programmatic replacement of structured expressions. The key thing to use this efficiently is that in ReplacePart you can use patterns to specify the coordinates of the things to be replaced, and you can define functions to apply to them.
Example with your data
ReplacePart[a, {_, 3} -> 0]
will replace every 3rd part of every sublist with 0.
ReplacePart[a, {i : _, 3} :> 2*a[[i, 3]]]
will double every 3rd part of every sublist.
As the authors suggest, the approaches based on Part need well-formed data, but Cases is built for robust separation of Lists:
Using your a,
a = {{{1, 2, 3}, {1, 3, 2}, 5}, {{1, 4, 5}, {1, 0, 2},
10}, {{4, 5, 3}, {8, 3, 4}, 15}};
Cases[a,{_List,_List,n_}:>n,Infinity]
{5, 10, 15}
The other pieces of a record can be extracted by similar forms.
Part-based approaches will gag on ill-formed data like:
badA = {{{1, 2, 3}, {1, 3, 2}, 5}, {{1, 4, 5}, {1, 0, 2},
10}, {{4, 5, 3}, {8, 3, 4}, 15}, {baddata}, {{1, 2, 3}, 4}};
badA[[All,3]]
{{{1, 2, 3}, {1, 3, 2}, 5}, {{1, 4, 5}, {1, 0, 2},
10}, {{4, 5, 3}, {8, 3, 4}, 15}, {baddata}, {{1, 2, 3},
4}}[[All, 3]]
,but Cases will skip over garbage, operating only on conforming data
Cases[badA, {_List, _List, s_} :> s, Infinity]
{5, 10, 15}
hth,
Fred Klingener
You can use Part (shorthand [[...]]) for this :
a[[All, 3]]
a[[All, {1, 2}]]
I am still not good working with lists in Mathematica the functional way. Here is a small problem that I'd like to ask what is a good functional way to solve.
I have say the following list made up of points. Hence each element is coordinates (x,y) of one point.
a = {{1, 2}, {3, 4}, {5, 6}}
I'd like to traverse this list, and every time I find a point whose y-coordinate is say > 3.5, I want to generate a complex conjugate point of it. At the end, I want to return a list of the points generated. So, in the above example, there are 2 points which will meet this condition. Hence the final list will have 5 points in it, the 3 original ones, and 2 complex conjugtes ones.
I tried this:
If[#[[2]] > 3.5, {#, {#[[1]], -#[[2]]}}, #] & /# a
but I get this
{{1, 2}, {{3, 4}, {3, -4}}, {{5, 6}, {5, -6}}}
You see the extra {} in the middle, around the points where I had to add a complex conjugate point. I'd like the result to be like this:
{{1, 2}, {3, 4}, {3, -4}, {5, 6}, {5, -6}}
I tried inserting Flatten, but did not work, So, I find myself sometimes going back to my old procedural way, and using things like Table and Do loop like this:
a = {{1, 2}, {3, 4}, {5, 6}}
result = {};
Do[
If[a[[i, 2]] > 3.5,
{
AppendTo[result, a[[i]]]; AppendTo[result, {a[[i, 1]], -a[[i, 2]]}]
},
AppendTo[result, a[[i]]]
],
{i, 1, Length[a]}
]
Which gives me what I want, but not functional solution, and I do not like it.
What would be the best functional way to solve such a list operation?
update 1
Using the same data above, let assume I want to make a calculation per each point as I traverse the list, and use this calculation in building the list. Let assume I want to find the Norm of the point (position vector), and use that to build a list, whose each element will now be {norm, point}. And follow the same logic as above. Hence, the only difference is that I am making an extra calculation at each step.
This is what I did using the solution provided:
a = {{1, 2}, {3, 4}, {5, 6}}
If[#[[2]] > 3.5,
Unevaluated#Sequence[ {Norm[#], #}, {Norm[#], {#[[1]], -#[[2]]}}],
{Norm[#], #}
] & /# a
Which gives what I want:
{ {Sqrt[5],{1,2}}, {5,{3,4}}, {5,{3,-4}}, {Sqrt[61],{5,6}}, {Sqrt[61],{5,-6}} }
The only issue I have with this, is that I am duplicating the call to Norm[#] for the same point in 3 places. Is there a way to do this without this duplication of computation?
This is how I currently do the above, again, using my old procedural way:
a = {{1, 2}, {3, 4}, {5, 6}}
result = {};
Do[
o = Norm[a[[i]]];
If[a[[i, 2]] > 3.5,
{
AppendTo[result, {o, a[[i]]}]; AppendTo[result, {o, {a[[i, 1]], -a[[i, 2]]}}]
},
AppendTo[result, {o, a[[i]]}]
],
{i, 1, Length[a]}
]
And I get the same result as the functional way, but in the above, since I used a temporary variable, I am doing the calculation one time per point.
Is this a place for things like sow and reap? I really never understood well these 2 functions. If not, how would you do this in functional way?
thanks
One way is to use Sequence.
Just a minor modification to your solution:
If[#1[[2]] > 3.5, Unevaluated#Sequence[#1, {#1[[1]], -#1[[2]]}], #1] & /# a
However, a plain ReplaceAll might be simpler:
a /. {x_, y_} /; y > 3.5 :> Sequence[{x, y}, {x, -y}]
This type of usage is the precise reason Rule and RuleDelayed have attribute SequenceHold.
Answer to update 1
I'd do it in two steps:
b = a /. {x_, y_} /; y > 3.5 :> Sequence[{x, y}, {x, -y}]
{Norm[#], #}& /# b
In a real calculation there's a chance you'd want to use the norm separately, so a Norm /# b might do
While Mathematica can simulate functional programming paradigms quite well, you might consider using Mathematica's native paradigm -- pattern matching:
a = {{1,2},{3,4},{5,6}}
b = a /. p:{x_, y_ /; y > 3.5} :> Sequence[p, {x, -y}]
You can then further transform the result to include the Norms:
c = Cases[b, p_ :> {Norm#p, p}]
There is no doubt that using Sequence to generate a very large list is not as efficient as, say, pre-allocating an array of the correct size and then updating it using element assignments. However, I usually prefer clarity of expression over such micro-optimization unless said optimization is measured to be crucial to my application.
Flatten takens a second argument that specifies the depth to which to flatten. Thus, you could also do the following.
a = {{1, 2}, {3, 4}, {5, 6}};
Flatten[If[#[[2]] > 3.5, {#, {#[[1]], -#[[2]]}}, {#}] & /# a, 1]
The most serious problem with your Do loop is the use of AppendTo. This will be very slow if result grows long. The standard way to deal with lists that grow as the result of a procedure like this is to use Reap and Sow. In this example, you can do something like so.
new = Reap[
Do[If[el[[2]] > 3.5, Sow[{el[[1]], -el[[2]]}]],
{el, a}]][[2, 1]];
Join[a, new]
To answer your edit, use With (or Module) if you're going to use something expensive more than once.
Here's my version of the problem in your edit:
a = {{1, 2}, {3, 4}, {5, 6}};
Table[With[{n = Norm[x]},
Unevaluated#Sequence[{n, x},
If[x[[2]] > 3.5, {n, {1, -1} x}, Unevaluated#Sequence[]]]],
{x, a}]
The structure of the above could be modified for use in a Map or ReplaceAll version, but I think that Table is clearer in this case. The unevaluated sequences are a little annoying. You could instead use some undefined function f then replace f with Sequence at the end.
Mark's Sow/Reap code does not return the elements in the order requested. This does:
a = {{1, 2}, {3, 4}, {5, 6}};
Reap[
If[Sow[#][[2]] > 3.5, Sow[# {1, -1}]] & /# a;
][[2, 1]]
You may use join with Apply(##):
Join ## ((If[#[[2]] > 3.5, {#, {#[[1]], -#[[2]]}}, {#}]) & /# a)
I have two list operations which I would like to ask for help. The way I implemented them is not very elegant, so I want to learn from you experts.
1) Suppose I have two lists, one is like {{0,2,4},{1,3,2},{2,0,4}} and the other is {{1,3,7},{2,4,6},{3,1,9}}. I want to either based on the value, or based on some criterion to filter through the first list, and then get the corresponding elements in the second. For example, based on value which is non-zero, I want to get {{3,7},{2,4,6},{3,9}}. Based on the condition greater than 2, I want to get {{7},{4},{9}}.
2) I have a list such as {{{1,2},{1,1}},{{1,3},{2,4}},{{1,2},{2,3}},{{1,4},{3,3}}}. I want to form {{{1,2},{{1,1},{2,3}}},{{1,3},{{2,4}}},{{1,4},{{3,3}}}. That is, I want to group those second lists if the first element is the same. How can I do this in a beautiful way?
Many thanks.
For the first part, you want Pick:
In[27]:= Pick[{{1,3,7},{2,4,6},{3,1,9}},{{0,2,4},{1,3,2},{2,0,4}},_?Positive]
Out[27]= {{3,7},{2,4,6},{3,9}}
In[28]:= Pick[{{1,3,7},{2,4,6},{3,1,9}},{{0,2,4},{1,3,2},{2,0,4}},_?(#>2&)]
Out[28]= {{7},{4},{9}}
For the second question, GatherBy gets you most of the way there:
In[29]:= x = GatherBy[{{{1, 2}, {1, 1}}, {{1, 3}, {2, 4}}, {{1, 2},
{2, 3}}, {{1, 4}, {3, 3}}}, First]
Out[29]= {{{{1, 2}, {1, 1}}, {{1, 2}, {2, 3}}}, {{{1, 3},
{2, 4}}}, {{{1, 4}, {3, 3}}}}
And then you can apply a rule to clean things up a bit:
In[30]:= x /. l:{{a_, _}..} :> {a, Last /# l}
Out[30]= {{{1, 2}, {{1, 1}, {2, 3}}}, {{1, 3}, {{2, 4}}},
{{1, 4}, {{3, 3}}}}
As Michael said, Pick is definitely the way to go for the first one.
For the second part, I'd like to offer an alternative that lets you do this in one line: SelectEquivalents. (I know, rather self promoting, but I use this function a lot.) To get the result your looking for, simply enter
In[1] := SelectEquivalents[ <list>, First, Last, {#1,#2}& ]
Out[1]:= {{{1, 2}, {{1, 1}, {2, 3}}}, {{1, 3}, {{2, 4}}}, {{1, 4}, {{3, 3}}}}
Internally, SelectEquivalents uses Reap and Sow, so First tags each element in <list>, Last transforms the element into the form we wish to use, and {#1, #2}& returns a list with elements of the form {Tag, {<items with that tag>}}. The advantage is that you get to specify everything in one step getting you what you want without subsequent transformations.