Plot a mean of ten lines of different resolution

Hello all,

I would like to plot the mean of ten lines. Each line plots the data from a different longitude and latitude against a timeslice. The data (mean annual temperature over the past 120,000 years) only covers temperatures on land, and as such, some timeslices have no corresponding temperature data for the years in which they were submerged.

The data dimensions are as follows:

Longitude: 720x1

Latitude: 300x1

Month: 12x1

Year: 72x1

Monthly temperature: 720x300x12x72

Can anyone advise on how to plot the mean of these ten timeslices? That is to say, the mean temperature over the past 120,000 years across the ten coordinate points plotted. I am very new to MATLAB and programming generally so help would be greatly appreciated.

my_year = -10000;

my_month = 6;

my_longitude = 11.2980;

my_latitude = 62.4342;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

[~,yearID] = min(abs(years - my_year));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#fff100');

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year');

ylabel('Temperature °C');

[daniel mitchell]

Could you clarify the specific assistance you need and describe the exact issue you're facing? Also, please provide the dimensions of your variables. This information may enable us to assist you more effectively.

[Robert]

Sorry, I am awfully new to MATLAB and programming/coding. I am trying to add an additional line to the plot showing the mean temperature for the geographic area selected (in the case of the attached plot these are 10 randomly selected coordinates between latitudes 60 and 65, and longitudes -10 and 50). I am going to repeat this for latitudes 55-60, 50-55, 45-50, 40-45, and 35-40, all within the same longitudes (this will cover Europe). I have edited the question to include the information you're asking about. I have also included that here:

Longitude: 720x1

Latitude: 300x1

Month: 12x1

Year: 72x1

Monthly temperature: 720x300x12x72

[MarKf]

Still unclear what you need the mean of and what do you need to plot from your description. What are the 10 lines in the 120,000 values?

I assume from the plot and the code that you want the annual average, that is averaging all the months in the years so to have a single mean value per year, to be clear, 72 yearly scores for each of the 720 longs times 300 lats. That would be 216000*72 average values, but I'm guessing some are missing so that's how you get 120000? (also misleading to call them 120,000 years, having 72 years of recording is impressive but hundreds of thousands? a strage way to put it).

Again from the plot, the missing values seem to be NaNs already, that'd be conveninent but do check that. In that case mean_annual = mean(temp,3,"omitnan") will ignore the missing values, otherwise just temp(indecesofmissing) = nan first, so that'd take care of that.

You're then finding an index of lats and longs (and also a yearID, but that's ignored) so I'm guessing that's how you choose those 10 annual vectors, there is no way for us to know without having the full code, variables and data. When you select those data, subsetemp = squeeze(mean_annual_temperature(lonID, latID, :) would work if the indeces lonID, latID were 2 ranges, so I assume the variables longitude and latitude are. And there's the rub for having "different resolutions of longitude and latitude against a timeslice", but again not sure what you need there.

[Robert]

Sorry, again new to MATLAB. The dataset covers 72 timeslices over the last 120,000 years from a palaeoclimate simulation (every 2,000 years except for the latest 22,000 years for which there is a resolution of every 1,000 years). The simulation has global coverage but as I am only seeking to analyse data for those parts of Europe not generally covered in icesheets I have limited the area to latitudes 35 to 65 and longitudes -10 to 50.

As temperatures would vary so much by topography even within similar latitudes, I elected to use a random coordinate generator (online) to output ten sets of coordinates per five degrees of latitude with the rule that none would be within 200km of eachother (see image). The image attached is a plot of the temperature for the ten randomly generated points over the 72 timeslices. What I wished to do was add a final line displaying a mean temperature for the entirety of that five degrees of latitude (though still between longitudes -10 and 50). I don't know how this could be done from the raw data as it requires coordinates to be inputted to produce temperature readings. Consequently, I assumed the mean would need to be the mean of the ten random sets of coordinates. I hope that makes sense. I know I am not describing it very well, but this really is all awfully new to me.

The code I provided in the question was the only code I have used, besides the ncread function to read in the data in the first place. I have since cut out some of the superfluous code such that it now reads as follows:

my_longitude = 31.7109;

my_latitude = 62.0342;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#fff100');

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year (10ka)');

ylabel('Temperature °C');

After this, I have been using the 'hold on' function and then been repeating the code for each set of coordinates viz.

hold on

my_longitude = 34.9305;

my_latitude = 64.9094;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#ff8c00');

Answer 1

[Star Strider]

It would help to have the actual data.

The missing data must be NaN values, since if they were simply blank, they would throw an error using the posted code. The result of ‘squeeze(mean_annual_temperature(lonID, latID, :))’ is therefore a 2D matrix, although I cannot determine its size relative to ‘years’ since plot corrects for that.

Guessing at the data, perhaps something like this —

mean_annual_temperature = randn(10, 72);                            % Create Data
mean_annual_temperature(5,[1:5 30:35 67:72]) = NaN;                 % Insert Some 'NaN' Values In One Row
years = 1:72;                                                       % Create Data

A = [years; mean_annual_temperature]                                % Concatenate MAtrices

A = 11×72

1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 19.0000 20.0000 21.0000 22.0000 23.0000 24.0000 25.0000 26.0000 27.0000 28.0000 29.0000 30.0000 -0.5411 -0.2815 1.0491 -1.6423 2.1098 1.6537 0.4762 2.0208 -0.5480 -0.3442 0.3323 -0.8276 -0.6295 1.0037 1.7795 1.2000 0.0774 1.6986 0.9332 0.4624 -0.1027 0.5142 -1.8523 -1.6349 -1.4737 0.6622 1.0846 0.1649 0.6325 -1.2057 -1.0054 0.6366 -0.9178 0.4337 0.6949 -1.7411 -0.3371 -1.3497 0.1201 -0.5295 0.2605 -0.3146 -0.2346 0.5556 0.7186 -0.9240 -0.1993 0.4710 1.4087 0.8780 -1.5684 0.6322 -1.2419 -0.7496 0.8837 -1.1482 2.2219 1.6758 -0.7045 -0.3722 0.1136 -0.0925 -0.1438 0.1713 0.4869 -0.1721 -1.9443 -0.2211 0.9828 -1.6542 0.6495 2.4467 -0.0475 -0.9288 -1.0449 1.5306 1.1406 0.7564 1.0304 0.5020 -0.9871 1.2289 0.9241 -0.6455 -0.1782 -1.1711 0.8850 0.3774 1.2468 0.7507 1.7575 1.5207 0.2616 -0.2608 -0.1276 0.9961 1.7219 0.1804 -0.2159 -0.7337 1.2809 1.3715 -0.0790 -0.2903 0.7787 1.3648 0.6253 0.2691 1.0305 0.9526 -0.1336 0.1418 -2.1785 -0.9253 -1.1874 -0.1016 0.1417 0.7380 -1.4048 -0.6412 NaN NaN NaN NaN NaN 2.5600 -0.4187 0.2515 -1.1774 1.1921 0.0762 -0.5691 0.2553 -0.7648 -0.7495 -0.5558 0.3327 -0.5510 0.1954 -0.3573 0.6896 1.5520 -1.9564 -1.5962 -1.4834 0.2716 0.9360 0.8312 -1.2839 NaN 0.7411 1.2922 0.2373 -0.6004 -1.4177 0.3602 0.8799 -1.1416 -0.8418 -0.0143 0.1472 -0.5124 1.2289 0.4280 -1.4072 -0.9609 1.3112 0.5484 0.8923 0.9473 0.2731 -1.2310 0.3174 -0.3024 -0.1986 2.2675 -0.5148 1.3039 1.3289 1.4054 0.0809 -0.4823 1.6888 -0.5762 0.1305 0.0925 0.6768 -0.3189 -0.8624 -1.2713 0.2825 -0.8371 -0.4500 -0.4663 -0.5002 -0.8410 0.0627 -0.7790 -0.1689 -0.8123 -0.1346 0.3743 0.3414 0.8541 1.0799 1.4315 0.5249 -0.1618 0.7157 -1.0191 0.7756 0.3965 -0.0901 -1.9013 -0.0282 0.4677 1.0814 -0.6354 -0.6108 1.2923 0.1929 -0.2663 -0.3312 -0.3346 -1.3464 -0.6880 0.9257 0.5566 -0.3718 0.1646 -0.1353 -0.4565 0.5285 0.1605 0.0936 -0.7124 -0.7214 -0.6197 0.0074 -1.3784 0.4037 0.0719 1.1228 -1.2194 0.6413 -0.7957 0.3585 0.2729 -0.8332 0.8570 2.8250 0.6536 0.1170 1.7289 0.9317 -0.5336 -1.7100 0.0384 0.7644 0.7899 1.0520 -1.1361 -1.6468 -1.2941 1.8891 0.7357 0.4758 -0.8837 0.2715 -1.3944

NaNidx = any(isnan(A),1)                                            % Determine 'NaN' Columns

NaNidx = 1×72 logical array
  Columns 1 through 49

   1   1   1   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   1   1   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0

  Columns 50 through 72

   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   1   1   1   1

A = rmmissing(A, 2)                                                 % Remove Missing Values Columns

A = 11×55

6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 19.0000 20.0000 21.0000 22.0000 23.0000 24.0000 25.0000 26.0000 27.0000 28.0000 29.0000 36.0000 37.0000 38.0000 39.0000 40.0000 41.0000 1.6537 0.4762 2.0208 -0.5480 -0.3442 0.3323 -0.8276 -0.6295 1.0037 1.7795 1.2000 0.0774 1.6986 0.9332 0.4624 -0.1027 0.5142 -1.8523 -1.6349 -1.4737 0.6622 1.0846 0.1649 0.6325 -0.6346 -1.8883 -0.9102 -0.7969 -2.7951 -1.1729 -1.7411 -0.3371 -1.3497 0.1201 -0.5295 0.2605 -0.3146 -0.2346 0.5556 0.7186 -0.9240 -0.1993 0.4710 1.4087 0.8780 -1.5684 0.6322 -1.2419 -0.7496 0.8837 -1.1482 2.2219 1.6758 -0.7045 -1.2104 -1.1930 -1.1528 -0.5900 0.3916 -2.2763 -0.1721 -1.9443 -0.2211 0.9828 -1.6542 0.6495 2.4467 -0.0475 -0.9288 -1.0449 1.5306 1.1406 0.7564 1.0304 0.5020 -0.9871 1.2289 0.9241 -0.6455 -0.1782 -1.1711 0.8850 0.3774 1.2468 -1.7037 1.4887 1.4091 0.2819 0.7183 -0.2286 0.9961 1.7219 0.1804 -0.2159 -0.7337 1.2809 1.3715 -0.0790 -0.2903 0.7787 1.3648 0.6253 0.2691 1.0305 0.9526 -0.1336 0.1418 -2.1785 -0.9253 -1.1874 -0.1016 0.1417 0.7380 -1.4048 0.6505 0.7621 -1.5573 0.8716 1.9005 2.3778 2.5600 -0.4187 0.2515 -1.1774 1.1921 0.0762 -0.5691 0.2553 -0.7648 -0.7495 -0.5558 0.3327 -0.5510 0.1954 -0.3573 0.6896 1.5520 -1.9564 -1.5962 -1.4834 0.2716 0.9360 0.8312 -1.2839 0.0489 -0.2204 1.1651 0.9799 -1.0178 1.3577 0.3602 0.8799 -1.1416 -0.8418 -0.0143 0.1472 -0.5124 1.2289 0.4280 -1.4072 -0.9609 1.3112 0.5484 0.8923 0.9473 0.2731 -1.2310 0.3174 -0.3024 -0.1986 2.2675 -0.5148 1.3039 1.3289 -0.9200 2.8640 0.1464 -0.4054 -1.5230 0.2206 0.0925 0.6768 -0.3189 -0.8624 -1.2713 0.2825 -0.8371 -0.4500 -0.4663 -0.5002 -0.8410 0.0627 -0.7790 -0.1689 -0.8123 -0.1346 0.3743 0.3414 0.8541 1.0799 1.4315 0.5249 -0.1618 0.7157 0.5353 1.1854 1.0517 0.2672 -0.0787 2.1001 0.4677 1.0814 -0.6354 -0.6108 1.2923 0.1929 -0.2663 -0.3312 -0.3346 -1.3464 -0.6880 0.9257 0.5566 -0.3718 0.1646 -0.1353 -0.4565 0.5285 0.1605 0.0936 -0.7124 -0.7214 -0.6197 0.0074 -1.7194 1.3012 0.1291 -1.3113 0.0739 0.3585 -0.7957 0.3585 0.2729 -0.8332 0.8570 2.8250 0.6536 0.1170 1.7289 0.9317 -0.5336 -1.7100 0.0384 0.7644 0.7899 1.0520 -1.1361 -1.6468 -1.2941 1.8891 0.7357 0.4758 -0.8837 0.2715 -0.4024 1.2618 -0.0973 0.2869 0.4551 -0.0912

Ayears = A(1,:)                                                     % Recover Vector

Ayears = 1×55

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 36 37 38 39 40 41

Amean = mean(A(2:end,:))                                            % Mean Of Remaining Columns

Amean = 1×55

0.3939 0.1929 -0.1931 -0.2693 -0.2191 0.3811 0.2106 0.1184 0.0441 -0.0865 -0.1509 0.2900 0.3111 0.6957 0.1766 -0.2510 0.1573 -0.7657 -0.6470 -0.0800 0.3188 0.5565 0.4261 0.0517 -0.4467 0.5210 0.0602 -0.1574 -0.0955 0.1871

B = NaN(size(years));                                               % Create 'NaN' Vector
B(~NaNidx) = Amean                                                  % Assign Mean Values To 'Non-NaN' Index Positions

B = 1×72

NaN NaN NaN NaN NaN 0.3939 0.1929 -0.1931 -0.2693 -0.2191 0.3811 0.2106 0.1184 0.0441 -0.0865 -0.1509 0.2900 0.3111 0.6957 0.1766 -0.2510 0.1573 -0.7657 -0.6470 -0.0800 0.3188 0.5565 0.4261 0.0517 NaN

figure
hp1 = plot(years, mean_annual_temperature, 'DisplayName','Temperature MAtrix');
hold on
hp2 = plot(years, B, '-k', 'LineWidth',2, 'DisplayName','Mean Temperatures');
hold off
title(['Mean annual temperature time series, 60N to 65N']);
xlabel('Year');
ylabel('Temperature °C');
legend([hp1(1) hp2], 'Location','best')

This plots only the data for the years in which no data are missing. I believe this is preferable to using the mean 'omitmissing' flag, since this approach only uses data from years in which all data are present.

.

[Robert]

The data comes from a journal paper found here: https://figshare.com/articles/dataset/LateQuaternary_Environment_nc/12293345/4

The relevant file is LateQuaternary_Environment.nc

The code I have been using is one I have edited slightly from that provided by the article's authors, which can also be downloaded from the same page in the file Matlab_script.m

I had never done anything with coding until the past few days which I am sure is telling to the generous people on this forum who have responded to my enquiry. The result of this is, however, that I really don't much understand how to go about doing what I am trying to do, nor the advice I am being given. Ordinarily, I would seek training in the software, but this is not feasible at the timescales I am working under for this particular project.

[Star Strider]

Thank you! That¹s a 1.24GB file, so this could take a few minutes. I’ll need to work with it offline, since I obviously can’t work with it here.

It would help to have your full code, so I understand how you’re reading and pre-processing the data. It would save me time, since I wouldn’t have to experiment to see how best to re-create your results.

I’m downloading the file now, and will wait to hear from you on the rest before proceeding.

EDIT — (9 Jul 2023 at 21:47)

Using this code:

clear; clc; close all;

file = 'LateQuaternary_Environment.nc';

%% Read variables from NetCDF file

longitude = ncread(file, 'longitude');

latitude = ncread(file, 'latitude');

years = ncread(file, 'time');

months = ncread(file, 'month');

temperature = ncread(file, 'temperature');

biome = ncread(file, 'biome');

%% Specify relevant point in time and space

my_year = -10000; % 10,000 BP

my_month = 6; % June

my_longitude = 0.1218;

my_latitude = 52.2053; % Cambridge, UK

figure(1);

%% Global biome map

subplot(2,2,1);

imagesc(longitude, latitude, biome(:,:, years == my_year)'); axis xy;

title(['Biome distribution, 10000 BP']);

%% Global temperature map

subplot(2,2,2);

imagesc(longitude, latitude, temperature(:,:, months == my_month, years == my_year)'); axis xy;

colorbar;

title(['Mean June temperature, 10000 BP']);

% use nearest grid points (alternatively, use 3D interpolation)

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

[~,yearID] = min(abs(years - my_year));

%% Monthly temperature distribution

subplot(2,2,3);

plot(months, squeeze(temperature(lonID, latID, :, yearID)), '-o');

title(['Monthly temperature distribution, Cambridge (UK), 10000 BP']);

xticks(1:12); xlabel("Month");

ylabel('Mean temperature');

%% Mean annual temperature time series

mean_annual_temperature = mean(temperature,3);

subplot(2,2,4);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), '-o');

title(['Mean annual temperature time series, Cambridge (UK)']);

xlabel('Year');

and in addition:

my_longitude = 31.7109;

my_latitude = 62.0342;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

MATMtx = squeeze(mean_annual_temperature(lonID, latID, :));

SzMATMtx = size(MATMtx);

figure

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#fff100');

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year (10ka)')

ylabel('Temperature °C')

I’m getting one line, not an array of them, as in the plot image in your original post.

What are you doing to get that matrix?

.

[Robert]

Thank you. The code I used to get the plot (though probably a very inefficient one) was:

file = 'LateQuaternary_Environment.nc';

longitude = ncread(file, 'longitude');

latitude = ncread(file, 'latitude');

years = ncread(file, 'time');

months = ncread(file, 'month');

temperature = ncread(file, 'temperature');

biome = ncread(file, 'biome');

my_longitude = 31.7109;

my_latitude = 62.0342;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#fff100');

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year (10ka)');

ylabel('Temperature °C');

hold on

my_longitude = 34.9305;

my_latitude = 64.9094;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#ff8c00');

hold on

my_longitude = 44.5189;

my_latitude = 62.2910;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#e81123');

hold on

my_longitude = 41.3837;

my_latitude = 64.4551;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#ec008c');

hold on

my_longitude = 5.0812;

my_latitude = 61.4402;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#68217a');

hold on

my_longitude = 40.5623;

my_latitude = 62.6844;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#00188f');

hold on

my_longitude = -1.6756;

my_latitude = 60.2894;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#00bcf2');

hold on

my_longitude = 26.6348;

my_latitude = 62.8649;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#00b294');

hold on

my_longitude = 11.2980;

my_latitude = 62.4342;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#009e49');

hold on

my_longitude = 24.5866;

my_latitude = 64.4031;

[~,lonID] = min(abs(longitude - my_longitude));

[~,latID] = min(abs(latitude - my_latitude));

mean_annual_temperature = mean(temperature,3);

plot(years, squeeze(mean_annual_temperature(lonID, latID, :)), 'Marker', 'none', 'Color', '#bad80a');

I should probably say that the coordinates used were generated outside MATLAB. Thinking about it I could make a variable containing the coordinate data (I currently have it in an excel spreadsheet), which would at least save me manually inputting the coordinates each time. I am sure this is an awfully inefficient code. I think I was just relieved that it worked to produce the following plot:

[Star Strider]

The code can be made a bit more efficient:

file = 'LateQuaternary_Environment.nc';

longitude = ncread(file, 'longitude');

latitude = ncread(file, 'latitude');

years = ncread(file, 'time');

months = ncread(file, 'month');

temperature = ncread(file, 'temperature');

biome = ncread(file, 'biome');

mean_annual_temperature = mean(temperature,3);

my_longitudev = [31.7109; 34.9305; 44.5189; 41.3837; 5.0812; 40.5623; -1.6756; 26.6348; 11.2980; 24.5866;];

my_latitudev = [62.0342; 64.9094; 62.2910; 64.4551; 61.4402; 62.6844; 60.2894; 62.8649; 62.4342; 64.4031;];

MAT1 = zeros(numel(my_longitudev), numel(years)); % Preallocate

for k = 1:numel(my_longitudev)

[~,lonID] = min(abs(longitude - my_longitudev(k)));

[~,latID] = min(abs(latitude - my_latitudev(k)));

MAT1(k,:) = squeeze(mean_annual_temperature(lonID, latID, :));

NrNaNs(k) = nnz(isnan(MAT1(k,:)));

end

MAT2 = MAT1; % Copy Original Matrix

NaNidx = any(isnan(MAT2),1); % Determine 'NaN' Columns

MAT2 = rmmissing(MAT2, 2); % Remove Missing Values Columns

MAT2years = years(~NaNidx); % Recover Vector

MAT2mean = mean(MAT2); % Mean Of Remaining Columns

MAT3 = NaN(size(years)); % Create 'NaN' Vector

MAT3(~NaNidx) = MAT2mean; % Assign Mean Values To 'Non-NaN' Index Positions

figure

hp1 = plot(years, MAT1, 'DisplayName','Temperature Matrix');

hold on

hp2 = plot(years, MAT3, '-k', 'LineWidth',2, 'DisplayName','Mean Temperatures');

hold off

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year');

ylabel('Temperature °C');

legend([hp1(1) hp2], 'Location','best')

Producing this plot —

The mean is calculated only over the complete rows. The ‘MAT2years’ is a vector of the years that are plotted, eliminating the columns with NaN values.

.

[Robert]

Thank you so much for all your help and patience. That code worked perfectly. Just one quick question before I accept the answer, if I may. How would I go about changing the line colours with the new code? I tried making a new colormap, though to no effect. I simply added the following code to the end of yours:

hold on

mycolors = [0.255 0.241 0.000; 0.255 0.140 0.000; 0.232 0.017 0.035; 0.236 0.000 0.140; 0.104 0.033 0.122; 0.000 0.024 0.143; 0.000 0.188 0.242; 0.000 0.178 0.148; 0.000 0.158 0.073; 0.186 0.216 0.010; 0.000 0.000 0.00];

ColorOrder = mycolors;

[Star Strider]

My pleasure!

The colororder call is incorrect.

It should be:

figure

hp1 = plot(years, MAT1, 'DisplayName','Temperature Matrix');

mycolors = [0.255 0.241 0.000; 0.255 0.140 0.000; 0.232 0.017 0.035; 0.236 0.000 0.140; 0.104 0.033 0.122; 0.000 0.024 0.143; 0.000 0.188 0.242; 0.000 0.178 0.148; 0.000 0.158 0.073; 0.186 0.216 0.010; 0.000 0.000 0.00];

colororder(mycolors)

hold on

hp2 = plot(years, MAT3, '-r', 'LineWidth',2, 'DisplayName','Mean Temperatures');

hold off

title(['Mean annual temperature time series, 60N to 65N']);

xlabel('Year');

ylabel('Temperature °C');

legend([hp1(1) hp2], 'Location','best')

although ‘mycolors’ can be assigned anywhere prior to the colororder call. That works as desired.

Also, an equivalent way of calculating the ‘MAT2’ mean is simply:

MAT2 = MAT1; % Copy Original Matrix

MAT3 = mean(MAT2);

since the NaN values produce NaN results for the mean in those columns.

My earlier code works correctly, and offers the option of doing different operations with the matrix, if those are contemplated. The rest of my code, other than the changes in this Comment, remains the same.

I changed the colour of the mean line to red because the other line colours are now all darker than in the original, and having it as black makes it difficult to distinguish. You can choose any colour for it that you want.

.

[Robert]

Thank you. For some reason they are all coming out black for me. I have, however, been enough of a nuisance to you I'm sure so I really don't think I can ask you to keep assisting me. Thanks ever so much for all your help.

[Star Strider]

My pleasure!

The colours are as you defined them.

To see that the idea actually works experiment with:

mycolors = colormap(turbo(10));

If my Answer helped you solve your problem, please Accept it!

.

Answer 2

[MarKf]

I see, this is overly complicated. Not as in complex or difficult, as in made needlessly overly complicated.

Let's just say you have 72 timepoints (or slices). You average the months as above (which is a point that is now unclear to me since we are talking about geological eras but whatever) and get 72 timepoints. I can see you already can plot those timepoints with the right scale with plot(years, mean_annual) so no probl. I'd say you'd be more interested in having the mean data points even if some locations or times are missing, so likely just 'omitnan' as I mention previously for now, rather than rmmissing as in the answer above.

In the map image you then have 10 random coordinates for each of 6 latitude ranges of 5 degrees, from south to north EU. So in this case you have 60 locations. Each location has a lat and a longit obvs. You can have a matrix of 60*2 of lats and a longits keeping track of the location and then a 60*72 mat of locations*timepoints vectors (timeseries). I'm unsure how you get the raw data then, but I assume you can easily create these 2 variable from what you got so far.

Then you just select the timeseries you want to average and/or plot based on the geograhical range of your choice. Maybe this code snippet can give you an idea.

lats = 35:5:65; longs = [-10 50]; lpoints = 10; 
yearsn = 72; years = [-120000:2000:-22000 -19000:1000:2000];
loc = [];
for il = 1:numel(lats)-1
    random_lats = (lats(il+1)-lats(il)).*rand(lpoints,1) + lats(il); %just a quickly randomly generated example
    random_longs = (longs(2)-longs(1)).*rand(lpoints,1) + longs(1); %likely not to have the 200km apart requirement
    loc = [loc;random_lats,random_longs]; %in this case they would be already ordered in groups of 5°
end
mean_annual_temp = randn(size(loc,1), yearsn); % random mean annual temperature data
mean_annual_temp(randi([1 numel(mean_annual_temp)],1,100)) = NaN; %add missing

%select locs based on lat (even if already ordered in groups of 5°, just in case)
range_lat = [43 48]; % you could also do this in a loop as above with  like for il = 1:numel(lats)-1, [lats(il) lats(il+1)]
loc_idx = loc(:,1)>=range_lat(1) & loc(:,1)<range_lat(2);
subsetmat = mean_annual_temp(loc_idx,:); % you could selct a time range with (:,time_idx); as well
subsetmean = mean(subsetmat,'omitnan');

%then same as above
figure
plot(years, subsetmat); %already plots all with diff colors
hold on
plot(years, subsetmean, '-k', 'LineWidth', 3);
hold off
title(sprintf('Mean annual temperature time series, %dN to %dN',range_lat));
xlabel('Year');
ylabel('Temperature °C');