假设我创建此表和值数组:
names = {'a'; 'b'; 'c'; 'd'} ; values = {'1'; '2'; '3'; '4'};
originalTable = table(names, values, 'VariableNames', {'names', 'values'});
nRepeat = [10, 50, 100, 2] ;
我想创建一个新表,其中将每行重复nRepeat对应索引的次数,即我将第一行或原始表重复10次,然后将原始表的第二行重复重复50次,依此类推... 此外,我想在该新表中添加一列带有重复索引。
我做了什么:
% Initialize newTable to allocate memory space
totalRepetitions = sum(nRepeat) ;
% Repeated first row of the original table the same number of times as the totalRepetitions that will happen, also adding the new column with the index of repetition
newTable = repmat([originalTable(1,:), array2table(1, 'VariableNames', {'idxRepetition'})], totalRepetitions , 1) ;
addedRows = 0 ;
for idxName = 1 : numel(originalTable.names)
newTable(addedRows +1 : addedRows + nRepeat(idxName) , :) =...
[repmat(originalTable(idxName ,:), nRepeat(idxName), 1), array2table( (1:1:nRepeat(idxName))', 'VariableNames', {'idxRepetition'}) ] ;
addedRows = addedRows + nRepeat(idxName);
end
这有效,但是对于大桌子却变得缓慢。
有没有更有效的方法?
答案 0 :(得分:7)
您只需在索引上使用repelem:
indx = repelem((1:numel(nRepeat)),nRepeat);
idxrep = arrayfun(@(x) 1:1:x,nRepeat,'un',0)'
finalTable = [originalTable(indx, :), table([idxrep{:}]','VariableNames', {'idxRepetition'})];
finalTable:
162×3 table
names values idxRepetition
_____ ______ _____________
'a' '1' 1
'a' '1' 2
'a' '1' 3
'a' '1' 4
'a' '1' 5
'a' '1' 6
'a' '1' 7
'a' '1' 8
'a' '1' 9
'a' '1' 10
'b' '2' 1
'b' '2' 2
'b' '2' 3
答案 1 :(得分:4)
这怎么样?
Irep = arrayfun(@(n) n*ones(1,nRepeat(n)), 1:length(nRepeat),'UniformOutput',false);
Irep = [Irep{:}]';
Iidx = arrayfun(@(n) 1:nRepeat(n), 1:length(nRepeat),'UniformOutput',false);
Iidx = [Iidx{:}]';
newTable = table(names(Irep),values(Irep), Iidx, 'VariableNames', {'names', 'values','idxRepetition'});
输出以下内容:
newTable =
162×3 table
names values idxRepetition
_____ ______ _____________
'a' '1' 1
'a' '1' 2
'a' '1' 3
'a' '1' 4
'a' '1' 5
'a' '1' 6
'a' '1' 7
'a' '1' 8
'a' '1' 9
'a' '1' 10
'b' '2' 1
'b' '2' 2
'b' '2' 3
'b' '2' 4
'b' '2' 5
...