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[?] Plot a table with sequence of dates?

Hi,

I am trying to draw a table like below. How can I make this work properly?

list = {{"Washington", "Washington", "Washington", "Washington", 
   "Washington", "Washington", "Washington"}, {1990.`, 1991.`, 1992.`,
    1993.`, 1994.`, 1995.`, 1996.`}, {"2", "3", "4", "5", "6", "7", 
   "8"}, {"400.3", "60.7", "46.8", "21.7", "16.4", "61.0", 
   "12.2"}, {"", "", "", "", "", "", ""}, {"Texas", "Texas", "Texas", 
   "Texas", "Texas", "Texas", "Texas"}, {1994.`, 1995.`, 1996.`, 
   1997.`, 1998.`, 1999.`, 2000.`}, {"1", "2", "3", "4", "5", "6", 
   "7"}, {"29.4", "58.6", "50.7", "56.4", "30.0", "37.5", "35.2"}}

enter image description here
POSTED BY: M.A. Ghorbani
21 Replies
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Another option is using NumberLinePlot

With[{opts={Ticks->None,Axes->False,AxesStyle->White,PlotRange->{{1990,2000},Automatic}}},
  Grid[
    {{"City","Date",""},
     {Item["Washington",Alignment->Left],"1990-1996",NumberLinePlot[Style[Interval[{1990,1996}],Red],opts]},
     {Item["Texas",Alignment->Left],"1994-2000",NumberLinePlot[Style[Interval[{1994,2000}],Darker@Green],opts]}},
    Dividers->{{None,Black,Black},{None,Black}}]]

grid
POSTED BY: Gustavo Delfino

Hi Rohit, thank you very much for this very interesting link! This is incredible! And it actually works:

enter image description here

Up to now I was always wondering how people can come up with undocumented functions ...

Best regards -- Henrik
POSTED BY: Henrik Schachner
Rohit Namjoshi
Rohit Namjoshi
Posted 6 years ago

Hi Henrik,

I think Interval only works with numeric values. Could use AbsoluteTime / FromAbsoluteTime to convert from / to DataObject. A Google search found this.
POSTED BY: Rohit Namjoshi

I guess you basically want TimelinePlot. If the diagram does not have to look exactly like the one shown (which is not very obvious anyway!), as a first try you can do like so:

list1 = Association[Rule @@@ Partition[Extract[list, {{1, 1}, {2}, {6, 1}, {7}}], 2]];
minmax = MinMax /@ list1;
list2 = Interval /@ Map[DateObject@*ToString, minmax, {2}];
tlp = TimelinePlot[list2, ImageSize -> {400, 100}];
Grid[{{TableForm[KeyValueMap[List, minmax], TableHeadings -> {None, {"City", "Date"}}, TableDepth -> 2], tlp}}, Frame -> All]

enter image description here

EDIT: Here a minor improvement (for the last line):

Grid[{{TableForm[
    KeyValueMap[
     List, #1 ~~ " - " ~~ #2 & @@@ 
      Map[StringDelete["."]@*ToString, minmax, {2}]], 
    TableHeadings -> {None, {"City", "Date"}}, TableDepth -> 2], 
   tlp}}, Frame -> All]

enter image description here
POSTED BY: Henrik Schachner

One more way: With

list1 = <|"Washington" -> {1966, 1967.`, 1968.`, 1969.`}, 
   "Texas" -> {1930, 1931.`, 1932.`, 1933.`, 1934.`, 1935.`}, 
   "Ohio" -> {1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970}|>;
minmax = MinMax /@ list1;
intv = Interval[{1963, 1969}];  (* the interval you are interested in *)
numlist = Interval /@ minmax;
intersect = Select[IntervalIntersection[#, intv] & /@ numlist, Length[#] != 0 &];

you can do:

Keys@intersect
(*  Out:  {"Washington","Ohio"}  *)
intersect["Ohio"]
(*  Out:  Interval[{1963,1969}]  *)

I found it surprising that I could not do any interval arithmetic with time intervals, e.g. this does not work:

IntervalIntersection[Interval[{DateObject[{1963}, "Year", "Gregorian", 1.`], DateObject[{1969}, "Year", "Gregorian", 1.`]}], 
 Interval[{DateObject[{1966}, "Year", "Gregorian", 1.`], DateObject[{1990}, "Year", "Gregorian", 1.`]}]]

Am I missing something?
POSTED BY: Henrik Schachner
Rohit Namjoshi
Rohit Namjoshi
Posted 6 years ago

Hi Mohammad,

Here is one way to do that

intervals = MinMax /* Round /* Interval /@ list1
intervals // Select[IntervalMemberQ[#, Interval[{1962, 1970}]] &] // Keys
(* {"Ohio"} *)

intervals // Select[IntervalMemberQ[Interval[{1962, 1970}], #] &] // Keys
(* {"Washington", "Ohio"} *)

To extract the values for a key

list1["Texas"]
(* {1930, 1931., 1932., 1933., 1934., 1935.} *)
POSTED BY: Rohit Namjoshi

Hello Mohammad,

For some reason your year date come as string - but they should be numbers. You can correct that like so:

list1 = Map[ToExpression, list1, {2}]
POSTED BY: Henrik Schachner
Rohit Namjoshi
Rohit Namjoshi
Posted 6 years ago

Hi Mohammad,

To remove the empty strings

list1 = Association[Rule @@@ Partition[Extract[m, {{1, 1}, {2}, {6, 1}, {7}}], 2]] // 
          Map[DeleteCases[#, ""] &]
POSTED BY: Rohit Namjoshi

@Gustavo Delfino: NumberLinePlot[Style[Interval[{...}] - Nice solution, I definitely like it!
POSTED BY: Henrik Schachner
Werner Geiger
Werner Geiger
Posted 6 years ago

Which data do you want to show in the third column?
POSTED BY: Werner Geiger

Hi Rohit,

I looked again. You are right absolutly. I really appreciate your help. Rohit, may I ask you about the last status Of complex network program? My friends and I look forward to see the incredible skill of you and the power of Mathematica in these kind of topics .

Thank you for your time.
POSTED BY: M.A. Ghorbani
Rohit Namjoshi
Rohit Namjoshi
Posted 6 years ago

Hi Mohammad,

The results look right to me. What were you expecting to see?
POSTED BY: Rohit Namjoshi

Hi Rohit,

Could you please have a look at the outputs for different intervals? 

intervals = <|"Washington" -> Interval[{1966, 2012}], 
   "Texas" -> Interval[{1970, 2014}], 
   "Ohio" -> Interval[{1930, 2015}], "Utah" -> Interval[{1966, 2010}],
    "Kansas" -> Interval[{1960, 2018}]|>;

intervals // 
  Select[IntervalMemberQ[#, Interval[{1966, 2012}]] &] // Keys

{"Washington", "Ohio", "Kansas"}

intervals // 
  Select[IntervalMemberQ[#, Interval[{1971, 2012}]] &] // Keys

{"Washington", "Texas", "Ohio", "Kansas"}

intervals // 
  Select[IntervalMemberQ[#, Interval[{1930, 2015}]] &] // Keys

{"Ohio"}
POSTED BY: M.A. Ghorbani

Dear Henrik and Rohit,

What kind of features can be extracted from the timelines? For example :

Which city's (cities') data is between 1962 and 1970? Answer: Ohio

How many cities have data from 1962 - 1970? Answer: Ohio and Washington.

Is it possible to extract the time series values for a specific city? Thank you.

list1 = <|"Washington" -> {1966, 1967.`, 1968.`, 1969.`}, 
  "Texas" -> {1930, 1931.`, 1932.`, 1933.`, 1934.`, 1935.`}, 
  "Ohio" -> {1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970}|>

minmax = MinMax /@ list1

list2 = Interval /@ Map[DateObject@*ToString, minmax, {2}]

tlp = TimelinePlot[list2, ImageSize -> {1000, 200}]
POSTED BY: M.A. Ghorbani

Why not simply

DeleteCases[#, ""] & /@ m

?
POSTED BY: Hans Dolhaine
POSTED BY: M.A. Ghorbani

Thank you so much Rohit.

Your help is always greatly appreciated
POSTED BY: M.A. Ghorbani

Hi Henrik,

Sometimes there are empty spaces on the list when I import the data from Excel.I couldn't delete these empty spaces using the DeleteCase[m," "] or DeleteCase[m,Null]. Is there any way to get rid of this problem?

In[22]:= m = {{"ACIPAYAM", "ACIPAYAM", "ACIPAYAM", "ACIPAYAM", "", 
    ""}, {"1966", "1966", "1966", "1966", "", ""}, {"2", "3", "4", 
    "5", "", ""}, {"19.5", "61.1", "17.2", "56.1", "", ""}, {"", 
    ""}, {"ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", 
    "ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", 
    "ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", "ADANA BÖLGE", "", 
    ""}, {"1930", "1930", "1930", "1930", "1930", "1930", "1930", 
    "1930", "1930", "1930", "1930", "1930", "", ""}, {"1", "2", "3", 
    "4", "5", "6", "7", "8", "9", "10", "11", "12", "", ""}, {"29.4", 
    "58.6", "50.7", "56.4", "30.0", "37.5", "35.2", "23.2", "33.0", 
    "49.2", "80.9", "188.2", "", ""}};

In[23]:= list1 = 
 Association[
  Rule @@@ Partition[Extract[m, {{1, 1}, {2}, {6, 1}, {7}}], 2]]

Out[23]= <|"ACIPAYAM" -> {"1966", "1966", "1966", "1966", "", ""}, 
 "ADANA BÖLGE" -> {"1930", "1930", "1930", "1930", "1930", "1930", 
   "1930", "1930", "1930", "1930", "1930", "1930", "", ""}|>

In[24]:= minmax = MinMax /@ list1

Out[24]= <|"ACIPAYAM" -> {Min["", "1966"], Max["", "1966"]}, 
 "ADANA BÖLGE" -> {Min["", "1930"], Max["", "1930"]}|>

In[25]:= list2 = Interval /@ Map[DateObject@*ToString, minmax, {2}]

Out[25]= <|"ACIPAYAM" -> 
  Interval[{DateObject[{1966}, "Year", "Gregorian", 3.5], 
    DateObject[{1966}, "Year", "Gregorian", 3.5]}], 
 "ADANA BÖLGE" -> 
  Interval[{DateObject[{1930}, "Year", "Gregorian", 3.5], 
    DateObject[{1930}, "Year", "Gregorian", 3.5]}]|>

In[26]:= tlp = TimelinePlot[list2, ImageSize -> {400, 100}]
POSTED BY: M.A. Ghorbani
Alex Teymouri
Alex Teymouri
Posted 6 years ago

A neat solution. Thank you Gustavo.
POSTED BY: Alex Teymouri

That is a great, Henrik.

i appreciate your time and consideration.
POSTED BY: M.A. Ghorbani

Hi Werner,

The first point is start of the date (1990) and the last point is end of the date(1996).

Thank you
POSTED BY: M.A. Ghorbani
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