How can I write a script to replace a specific integer value in my table, with the average of the data above and below that integer in the table

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JJ
JJ 2024-7-20
编辑： dpb 2024-7-21
Basically, I am trying to write a script to be able to find the value 999 in my table, and add the nearest possible values above and below the 999.
For example, if 999 were a value in row 10, column 4, I would want to add 12+18, then divide by 2, and then replace that 999. If 12, 16, 18 were all set to 999, I would want to add 10 and 22, then divide by 2. Then replace 12, 16, 18 all with 16.
Complete beginner to matlab, so I don't understand the syntax of how I would do this. Is this even possible with any sort of loop?
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sai charan sampara
sai charan sampara 2024-7-20
Ran in:
Hello JJ,
You can use logical indexing to find all the index values containing the value "999". You can replace these values with "NaN" values and then use "fillmissing" with "linear" method. In this case if 12,16,18 were all set to 999 then the values will be re written as 13,16,19 instead of all three being 16,16,16. If having all three as 16 is necessary, you can try something similar to the code below:
load patients
T = table(Age,Height,Weight,Systolic,Diastolic, ...
'RowNames',LastName);
T=T(1:20,:);
idx=[3,5,8,9,10,13,16,17];
T{idx,"Weight"}=999;% Generating example data
T
T = 20x5 table
Age Height Weight Systolic Diastolic ___ ______ ______ ________ _________ Smith 38 71 176 124 93 Johnson 43 69 163 109 77 Williams 38 64 999 125 83 Jones 40 67 133 117 75 Brown 49 64 999 122 80 Davis 46 68 142 121 70 Miller 33 64 142 130 88 Wilson 40 68 999 115 82 Moore 28 68 999 115 78 Taylor 31 66 999 118 86 Anderson 45 68 128 114 77 Thomas 42 66 137 115 68 Jackson 25 71 999 127 74 White 39 72 202 130 95 Harris 36 65 129 114 79 Martin 48 71 999 130 92
idx2=find(T{:,'Weight'}==999);
arr_new=T{:,'Weight'};
for i=1:length(idx2)
i_before=idx2(i)-1;
i_after=idx2(i)+1;
while(T{i_before,'Weight'}==999||T{i_after,'Weight'}==999)
if(T{i_before,'Weight'}==999)
i_before=i_before-1;
end
if(T{i_after,'Weight'}==999)
i_after=i_after+1;
end
end
arr_new(i_before+1:i_after-1)=(T{i_before,'Weight'}+T{i_after,'Weight'})*0.5.*ones(length(i_before+1:i_after-1),1);
end
T{:,"Weight"}=arr_new;
T
T = 20x5 table
Age Height Weight Systolic Diastolic ___ ______ ______ ________ _________ Smith 38 71 176 124 93 Johnson 43 69 163 109 77 Williams 38 64 148 125 83 Jones 40 67 133 117 75 Brown 49 64 137.5 122 80 Davis 46 68 142 121 70 Miller 33 64 142 130 88 Wilson 40 68 135 115 82 Moore 28 68 135 115 78 Taylor 31 66 135 118 86 Anderson 45 68 128 114 77 Thomas 42 66 137 115 68 Jackson 25 71 169.5 127 74 White 39 72 202 130 95 Harris 36 65 129 114 79 Martin 48 71 130 130 92
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Walter Roberson
Walter Roberson 2024-7-20
You can use logical indexing to find all the index values containing the value "999". You can replace these values with "NaN" values and then use "fillmissing" with "linear" method.
You do not happen to show that algorithm.
One way to replace the 999 with NaN is to use standardizeMissing

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dpb
dpb 2024-7-20
编辑：dpb 2024-7-21
Ran in:
"what function could make me fill it with the expected value?"
There isn't a builtin that will do that automagically, but if you save a lookup table of the pattern, then
HR_PAT=[0 4 6 10 12 16 18 22]; % the pattern
h=repmat(HR_PAT,1,5).'; % base vector
N=5; % number bad elements
isbad=sort(randperm(numel(h),N))
isbad = 1x5
7 14 16 17 37
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
h(isbad)=999; % the bad locations
h(isbad)=interp1(HR_PAT-0.1,HR_PAT,h(isbad-1),'next',0); % fill with the next; from previous, extrap --> 0
h
h = 40x1
0 4 6 10 12 16 18 22 0 4
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Observe the value 18 in the seventh location above that was set to 999 as bad element.
The "trick" is using the HR_PAT-0.1 vector as the x variable in the interp1 call; even with the 'next' method, interp1 returns the exact location if it finds an exact match as it always would if the X lookup values were the exact pattern values. To illustrate that (imo unfortunate) design choice observe
>> interp1(HR_PAT,HR_PAT,22,'next',0)
ans =
22.00
returned the same value as the lookup input value when is identical match.
To start over again, the value of a lookup for an input of 22 the next element is out of range so the 0 extrapolation value is returned.
>> interp1(HR_PAT-0.1,HR_PAT,22,'next',0)
ans =
0
returned the extrapolated value to begin the sequence over again. Note this is again using the "-0.1" X table values to ensure no exact match so the interpolant is the next table value.
The values to lookup are the location of the previous location prior to each bad value, hence h (isbad-1)

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