Extract data in a table following a a range of date (years)

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alejandro paradiso
alejandro paradiso 2022-11-18
编辑： Campion Loong 2022-11-21
Good day to all,
I have a thre column table, after some manipulation I get the years from dd/mm/yyyy to only yyyy. Now I have to filter this table using year as filter variable.
Date starts at 1900 and ends at 2050, the interval i need is from 1956 to 2020.
I have created a logical vector using
yearFiltered=newDataFiltered.yearFiltered(newDataFiltered.yearFiltered>1955 & newDataFiltered.yearFiltered<2021)
then i've tried to index the main data set using the variable above following A(B)=[]
the outcome of the syntax is that "when deleting elements from a table variable using indexed assignments, the number of the rows must not change. Specify the first subscript as a colon (:) and exactly one other subscript that is not a colon"
infact the rows number will be reduced.

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采纳的回答

Cris LaPierre
Cris LaPierre 2022-11-18
Your table has 3 columns. Your deletion code must delete all 3 columns. You specify 'all columns' by adding a colon in the 2nd position.
A(B,:)=[]
% ^^ add colun to second index position
  2 个评论
alejandro paradiso
alejandro paradiso 2022-11-18
thank you cris,
infact I want also erase the rows from the other columns
'Monday' '08:07:30 PM' 1900
'Wednesday' '02:50:12 PM' 1910
'Friday' '09:11:48 AM' 1920
'Sunday' '02:02:06 AM' 1930
'Monday' '04:36:36 PM' 1940
'Wednesday' '04:38:36 AM' 1950
'Thursday' '02:21:54 PM' 1900
from the data above, lets say i'm interested (columns & rows) in the years from 1920 to 1940, so in the end i will have a table 3x3
Cris LaPierre
Cris LaPierre 2022-11-18
Ran in:
Use a logical expression to identify the rows within the range of years you want, and then either extract those rows, all columns to a new variable or delete all other rows, all columns by taking the NOT of your logical expression.
a = ([1:3;4:6])'
a = 3×2
1 4 2 5 3 6
idx = a(:,1)==2
idx = 3×1 logical array
0 1 0
% Extract row
b = a(idx,:)
b = 1×2
2 5
% delete other rows
a(~idx,:)=[]
a = 1×2
2 5

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更多回答（1 个）

Campion Loong
Campion Loong 2022-11-18
编辑：Campion Loong 2022-11-21
Ran in:
I would use timetable for this kind of operations. It is specifically built for time based workflows.
% Mock data for time between 1900 - 2050
dt = datetime(1900,1,1)+calmonths(1:12*150)';
data = (1:length(dt))';
% timetable automatically picks up the one datetime variable as your time vector
tt = timetable(dt, data)
tt = 1800×1 timetable
dt data ___________ ____ 01-Feb-1900 1 01-Mar-1900 2 01-Apr-1900 3 01-May-1900 4 01-Jun-1900 5 01-Jul-1900 6 01-Aug-1900 7 01-Sep-1900 8 01-Oct-1900 9 01-Nov-1900 10 01-Dec-1900 11 01-Jan-1901 12 01-Feb-1901 13 01-Mar-1901 14 01-Apr-1901 15 01-May-1901 16
Then use timerange to select data you are interested in
% Make a timerange subscript for 1956 - 2020
tr = timerange(datetime(1956,1,1),datetime(2020,12,31))
tr =
timetable timerange subscript: Select timetable rows with times in the half-open interval: [01-Jan-1956 00:00:00, 31-Dec-2020 00:00:00) See Select Times in Timetable.
% Get all the rows within this range of time
tt(tr, :)
ans = 780×1 timetable
dt data ___________ ____ 01-Jan-1956 672 01-Feb-1956 673 01-Mar-1956 674 01-Apr-1956 675 01-May-1956 676 01-Jun-1956 677 01-Jul-1956 678 01-Aug-1956 679 01-Sep-1956 680 01-Oct-1956 681 01-Nov-1956 682 01-Dec-1956 683 01-Jan-1957 684 01-Feb-1957 685 01-Mar-1957 686 01-Apr-1957 687

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