の数字C
に割り当てる文字の配列を としますA
。それで
A = [1 3 5];
B = {[1]; [3]; [5]; [1;3]; [1;5]; [3;5]; [1;3;5]};
C = ['a', 'b', 'c']
k = 6; % indicates current line of B
str = ['you can have ' strrep(strrep(sprintf('_%c_', ...
C(ismember(A, B{k}))'), '__', ' or '), '_', '')];
結果は
str =
you can have a or b or c
一度にすべてのフィールドへの応答を作成する 場合は、次を使用できますB
allStr = arrayfun(@(x) ['you can have ' strrep(strrep(sprintf('_%c_', ...
C(ismember(A, B{x}))'), '__', ' or '), '_', '')], ...
(1:length(B))', 'uniformoutput', false)
これにより、
allStr =
'you can have a'
'you can have b'
'you can have c'
'you can have a or b'
'you can have a or c'
'you can have b or c'
'you can have a or b or c'
このコードの段階的な説明は次のとおりです。
% which contents of A can be found in B?
idx = ismember(A, B{k})';
% to which letters do these indices correspond?
letters = C(idx);
% group the letters in a string embedded in '_' as place holders for later use
% by this, the places between letters will be marked with '__' and the places
% at the beginning and the end of the string will be marked with '_'
stringRaw = sprintf('_%c_', letters);
% replace each occurrence of '__' by ' or '
stringOr = strrep(stringRaw, '__', ' or ');
% replace each occurrence of '_' by ''
stringClean = strrep(stringOr, '_', '');
% add first half of sentence
stringComplete = ['you can have ' stringClean];
(コメントで要求されているように)完全な単語でこれを機能さC
せるには、文字列のセル配列に変換し、それに応じて式を更新する必要があります。
A = [1 3 5];
B = {[1]; [3]; [5]; [1;3]; [1;5]; [3;5]; [1;3;5]};
C = {'first', 'second', 'third'}
k = 7; % indicates current line of B
str = ['you can have ' strrep(strrep(sprintf('_%s_', ...
C{ismember(A, B{k})}), '__', ' or '), '_', '')];
これにより、次の結果が得られます。
str =
you can have first or second or third