Question

在MATLAB中，我有一个这样的表：

//project text to ASCII Dec
#include <stdio.h>

main(){
    char i=0, data, a;

    printf("Enter data : ");
    scanf("%s", &data); //or %c

    while(data != i){ /* can't out from this loop I mean to the next letter of word */
      a = data;
      printf("%d", data); // or printf("%d", a);
      if(data != 0){
        printf("-");
      }
    }
}

如何将date foo bar mitzva baz 1995-1-1 21 54 12 37 1995-1-2 NaN 52 11 36 1995-1-3 39 56 NaN 38 1995-1-4 25 NaN NaN 50的任何列设置为其行NaN的值？

或换句话说：

如何设置过滤器，以便在baz之前的三列中的任何一列中逐行检查NaN？
如何设置三个值而不复制好值3次？

为了让它发挥作用，我在下面做了哪些改变？还是我的方向错了？

baz

我想得到：

colsToUpdate = {colidxFoo : colidxMitzva);
baddataFilter = any(isnan(data(:,colsToUpdate)); 
    % corrected from isnull. thanks @SardarUsama my mistake...

data(baddataFilter, colsToUpdate) = data(baddataFilter, colidxBaz);

Answer 1

Val = table2array(T(:,2:end));  %Extracting the elements of all columns except 1st one
IndBadV = isnan(Val);           %Finding NaNs
Val(IndBadV) = 0;               %Replacing NaNs with zeros
Val = Val+repmat(Val(:,end),1,size(Val,2)) .* IndBadV; %Updating the values
T = [T(:,1) array2table(Val, ...%Updating the table
    'VariableNames',T.Properties.VariableNames(2:end))];

Answer 2

只是为了澄清乌萨马的答案：

% problemCols = 2:end
Data = table2array(T(:,2:end));  % Without date column (which serves as index)
BadOnes = isnan(Data);           % Finding NaNs
Data(BadOnes) = 0;               % Replacing NaNs with zeros
FixcolsCount = size(Data,2);     % Number of columns to fix
GoodCol = end;
RepFixed = repmat(Data(:,GoodCol),1,FixcolsCount); % repeat 1 time down, for all columns
Data = Data + RepFixed .* BadOnes;       % Updating the values
T = [T(:,1) array2table(Val, ...         % Updating the table
    'VariableNames',T.Properties.VariableNames(2:end))];

注意：这是有效的，因为当数据存在时添加了零（badones为零），但是当数据不存在时添加了Repfix值（在这种情况下，badones为1）。

例如：

index  foo  bar mitzva baz     badones a   b   c      fixed  foo bar mitzva
1      11   na  13     14              0    1  0             0    14  0
2      na   22  23     24              1    0  0             24    0  0

加法结果和。*乘法：

1      11  14   13     14
2      24  22   23     24

一位同事用另一种方式告诉我：

BadOnes = isnan(T{:,2:end-1});          
GoodIndice = find(~BadOnes);
Fixed = BadOnes * repmat(T{:,end},1,size(T{:,2:end-1},2));;
Fixed(GoodIndice) = DataArray(GoodIndice); % adds GoodValues back into array
T{:,2:end-1} = Fixed;

或详细解释：

DataArray = T{:,2:end-1};
ProblemColsCount = size(DataArray,2); % Number of columns to fix

BadOnes = isnan(DataArray);           % Finding NaNs
% GoodCol = end;
RepFixed = repmat(T{:,end},1,ProblemColsCount); % repeat 1 time down, for all columns

GoodIndice = find(~BadOnes);
Fixed = BadOnes * RepFixed;
Fixed(GoodIndice) = DataArray(GoodIndice); % adds GoodValues back into array
T{:,2:end0-1} = Fixed;

在任何情况下，我都希望有一些像Usama的原始答案fillmissing()或类似的东西。

将任何特定列中的NaN替换为另一列中的值

2 个答案: