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[?] Plot a table with sequence of dates?

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

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"}}

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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M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

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

Posted 6 years ago

Hi Mohammad, The results look right to me. What were you expecting to see?

POSTED BY: Rohit Namjoshi

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

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"}

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

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

Hi Rohit, thank you very much for this very interesting link! This is incredible! And it actually works: Up to now I was always wondering how people can come up with undocumented functions ... Best regards -- Henrik

POSTED BY: Henrik Schachner

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

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

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?

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

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.} *)

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

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

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}]

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

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

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

Hans Dolhaine

Hans Dolhaine, retired

Posted 6 years ago

Why not simply DeleteCases[#, ""] & /@ m ?

POSTED BY: Hans Dolhaine

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

POSTED BY: M.A. Ghorbani

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

Thank you so much Rohit. Your help is always greatly appreciated

POSTED BY: M.A. Ghorbani

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

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

POSTED BY: M.A. Ghorbani

Alex Teymouri

Posted 6 years ago

A neat solution. Thank you Gustavo.

POSTED BY: Alex Teymouri

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

POSTED BY: Henrik Schachner

Gustavo Delfino

Posted 6 years ago

POSTED BY: Gustavo Delfino

M.A. Ghorbani

M.A. Ghorbani, University of Tabriz

Posted 6 years ago

That is a great, Henrik. i appreciate your time and consideration.

POSTED BY: M.A. Ghorbani

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

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] 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]

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