Concatenate Tables with/without "Join"

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Dario Walter
Dario Walter 2019-6-17
评论： Dario Walter 2019-6-18
Hey guys,
I could really need your help right now!
Let us assume
A = table(['A';'B';'C';'D';'E'],[38;43;38;40;49],...
'VariableNames',{'Character' 'Time1'})
B = table(['A';'C';'D';'E';'F'],[92;46;29;50;57],...
'VariableNames',{'Character' 'Time2'})
I want to create table in such a way that it returns
C=table(['A';'B';'C';'D';'E';'F'],[38;43,38;40;49;nan],[92;nan;46;29;50;57])
Do you have any idea? Thanks for your help!
  2 个评论
Matt J
Matt J 2019-6-17
Your code to generate C generates an error
>> C=table(['A';'B';'C';'D';'E';'F'],[38;43,38;40;49;nan],[92;nan;46;29;50;57])
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
And what do you mean by "with/without Join"? Do you mean either is acceptable?
Dario Walter
Dario Walter 2019-6-17
Please apologize, there was a comma instead of a semicolon.
C=table(['A';'B';'C';'D';'E';'F'],[38;43;38;40;49;nan],[92;nan;46;29;50;57])
Exactly, I do not mind how it works. It just has work.

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采纳的回答

Matt J
Matt J 2019-6-17
编辑：Matt J 2019-6-17
>> C = outerjoin(A,B,'Keys','Character','MergeKeys',true)
C =
6×3 table
Character Time1 Time2
_________ _____ _____
A 38 92
B 43 NaN
C 38 46
D 40 29
E 49 50
F NaN 57

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