Simple but puzzling plotting question

Hi,

You can do this only if each of the data sets use the same x-coordinate.

l1 = Table[{x, Sin[Pi x]}, {x, -Pi, Pi, Pi/8}];
l2 = Table[{x, Cos[Pi x]}, {x, -Pi, 0, Pi/12}];
l3 = Table[{x, Tan[Pi x]}, {x, 0, Pi, Pi/16}];
ListPlot[{l1, l2, l3}, Joined -
In your case, you'd want to plot each of your currencies as they vary with
date.

Regards,

Dave.

If you want the data correctly aligned on the right hand side and no
annoying zeros on the left, one can use an index (or rescale an existing
index: that all depends on the actual data structure you are using) as
in the following example:

data1 = {54.8, 54.6, 54.78, 56.19, 56.97, 57.16, 57.7, 57.72, 56.28,
56.78, 56.63, 55.77, 54.92, 55.6, 54.86, 53.6, 53.54, 53.49, 53.39,
51.59, 53.2, 55.09};
data2 = {89.89, 89.01, 89.39, 88.89, 88.18, 87.41, 86.84, 87.07, 87.06,
88.05, 88.22, 88.55, 88.13, 88.2, 88.89};

data1 = Table[{i, data1[[i]]}, {i, Length[data1]}];
data2 = Table[{i + Length[data1] - Length[data2], data2[[i]]},
{i, Length[data2]}];

<< "Graphics`MultipleListPlot`"
MultipleListPlot[{data1, data2}]

Regards,
Jean-Marc

Using version 6, this can be done easily as follows:

First generate two data sets of differing length:

set1 = Sort@Table[RandomReal[], {5}];
set2 = Sort@Table[RandomReal[], {10}];

then plot them:

ListPlot[{set1, set2}, Joined -
Or if you are using version 5.x, you can get a similar result with:

set1 = Sort@Table[Random[], {5}];
set2 = Sort@Table[Random[], {10}];

Show[
Block[{$DisplayFunction=Identity},
{ListPlot[set1, PlotJoined- ListPlot[set2, PlotJoined- PlotRange- --
To reply via email subtract one hundred and four

In version 5.2, you could use *MultipleListPlot* and *PadLeft* with 0
the missing data to get lists of equal length. (Note that
*MultipleListPlot* will plot lists of different length but the smaller
list will be padded on the right. See below.) For instance,

In[1]:=
data1 = {54.8, 54.6, 54.78, 56.19, 56.97, 57.16,
57.7, 57.72, 56.28, 56.78, 56.63, 55.77, 54.92,
55.6, 54.86, 53.6, 53.54, 53.49, 53.39, 51.59,
53.2, 55.09};
data2 = {89.89, 89.01, 89.39, 88.89, 88.18, 87.41,
86.84, 87.07, 87.06, 88.05, 88.22, 88.55, 88.13,
88.2, 88.89};
Length /@ {data1, data2}

Out[3]=
{22, 15}

In[4]:=
<< "Graphics`MultipleListPlot`"
MultipleListPlot[{data1, data2}]
data3 = PadLeft[data2, Length[data1]];
MultipleListPlot[{data1, data3}]

Regards,
Jean-Marc

In version 6.0, *DateListPlot* manages very well to plot lists of
various sizes, as you can see in the following example.(A demo notebook
and a PDF file are available at
#
#
respectively.)

In[1]:= ms = FinancialData["MS", {"March 1, 2005", "March 31, 2005"}]
ibm = FinancialData["IBM", {"March 10, 2005", "March 31, 2005"}]
Length /@ {ms, ibm}
DateListPlot[{ms, ibm}, Joined -
Out[1]= {{{2005, 3, 1}, 54.8}, {{2005, 3, 2}, 54.6}, {{2005, 3, 3},
54.78}, {{2005, 3, 4}, 56.19}, {{2005, 3, 7}, 56.97}, {{2005, 3, 8},
57.16}, {{2005, 3, 9}, 57.7}, {{2005, 3, 10},
57.72}, {{2005, 3, 11}, 56.28}, {{2005, 3, 14},
56.78}, {{2005, 3, 15}, 56.63}, {{2005, 3, 16},
55.77}, {{2005, 3, 17}, 54.92}, {{2005, 3, 18},
55.6}, {{2005, 3, 21}, 54.86}, {{2005, 3, 22},
53.6}, {{2005, 3, 23}, 53.54}, {{2005, 3, 24},
53.49}, {{2005, 3, 28}, 53.39}, {{2005, 3, 29},
51.59}, {{2005, 3, 30}, 53.2}, {{2005, 3, 31}, 55.09}}

Out[2]= {{{2005, 3, 10}, 89.89}, {{2005, 3, 11},
89.01}, {{2005, 3, 14}, 89.39}, {{2005, 3, 15},
88.89}, {{2005, 3, 16}, 88.18}, {{2005, 3, 17},
87.41}, {{2005, 3, 18}, 86.84}, {{2005, 3, 21},
87.07}, {{2005, 3, 22}, 87.06}, {{2005, 3, 23},
88.05}, {{2005, 3, 24}, 88.22}, {{2005, 3, 28},
88.55}, {{2005, 3, 29}, 88.13}, {{2005, 3, 30},
88.2}, {{2005, 3, 31}, 88.89}}

Out[3]= {22, 15}

Out[4]= [... plot deleted ...]

Regards,
Jean-Marc