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How to join the subelements of a cell?

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GEORGIOS BEKAS
GEORGIOS BEKAS 2018-1-26
评论： Domanic 2018-1-26
I am trying to express all sentences in one cell with string subelements.
s='I am George. John is there? Ana is around! Are you mad?'
[y,matches] = strsplit(s,{'.','?','!'})
for i = 1:length(matches)
y{i}=[y{i},matches{i}]
end
y=y(~cellfun('isempty',y))
  1 个评论
Matt J
Matt J 2018-1-26
Don't think you finished your question (because I'm not sure what it is). If you're trying to rejoin all the sub-strings in y, see my answer below.

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回答（2 个）

Matt J
Matt J 2018-1-26
编辑：Matt J 2018-1-26
s=[y{:}]
  6 个评论
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GEORGIOS BEKAS
GEORGIOS BEKAS 2018-1-26
seperate strings into one cell subelement.
Steven Lord
Steven Lord 2018-1-26
This is what I receive when I run the code you posted followed by Matt's suggestion.
s = 'I am George. John is there? Ana is around! Are you mad?'
[y,matches] = strsplit(s,{'.','?','!'})
for i = 1:length(matches)
y{i}=[y{i},matches{i}]
end
y=y(~cellfun('isempty',y))
s2 = [y{:}]
whos s2
s2 =
'I am George. John is there? Ana is around! Are you mad?'
Name Size Bytes Class Attributes
s2 1x55 110 char
What exactly do you need/want/expect s2 to be for this example? Be specific about the type and size you expect s2 to be and the contents you expect s2 to contain.

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Domanic
Domanic 2018-1-26
编辑：Domanic 2018-1-26
To join elements of the cell array, y, you can use:
y = cell2mat(y);
or
y = {cell2mat(y)};
for a cell array.
To more generally achieve what you're trying to do, you can use regular expressions:
expression = '(?<=\W)\s'; % Expression to find white space following punctuation
strsplit(s,expression,'delimitertype','regularexpression'); % split string around result of regexp
  2 个评论
GEORGIOS BEKAS
GEORGIOS BEKAS 2018-1-26
the problem is that the strings are not separate. C[1,1] = 'I am George.' 'John is there?' 'Ana is around!' 'Are you mad?'
Domanic
Domanic 2018-1-26
How about:
expression = '(?<=\W)\s';
y = strsplit(s,expression,'delimitertype','regularexpression');
y = {cell2struct(y,'substrings')};
where the strings are accessed as:
y{1}.substrings
or
y{1}.substrings(1)
not sure why you'd want this, though.

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