Error using containers.Map

15 views (last 30 days)
jungdnb
jungdnb on 30 Nov 2016
Commented: Walter Roberson on 3 Nov 2017
Hello, I am having following problem. I want to display single cells from my string array if my function finds a specific string. Here is the code:
function voddi_test(graph)
global vodcode
system_name = extract_system_name(graph);
variables = containers.Map();
variable = containers.Map();
keys = variables.keys;
for name = variables.keys
variable(name{1}) = true;
end
for node_name = filter_graph(graph, 'Variable', true)
node = graph(node_name{1});
for arg = node.arguments
variables([system_name '/' arg{1}]) = true;
end
end
for keys = 1:numel(variables)
if strcmp(variables(keys), 'Constant')
variables(keys{1})
else
variables(keys{2})
end
end
end
I am keeping getting the error message "Specified key type does not match the type expected for this container" when I try compare the content of the variables map with my desired strings. The code works fine, expect for the last for - loop. Any help is much appreciated, many thanks.

Accepted Answer

Guillaume
Guillaume on 1 Dec 2016
Note: I would avoid having variable names that only differ by one being plural and singular unless one was just a scalar element of the plural container.
As per Walter's answer, it seems you have not shown all the code, since your first loop is useless if the map is not filled.
Assuming the map declaration you've shown is the actual you use:
variables = containers.Map();
then it uses 'char' as the key type. However, you then have:
for keys = 1:numel(variables)
if strcmp(variables(keys), 'Constant')
There are two issues here. For starter, since variables is a map, numel(variables) is just 1. Perhaps you meant, variables.Count. Secondly, keys will be a double. As said the key type is 'char', not 'double', so you can't actually use your keys as a key into the map. Hence the error message.
Perhaphs, you meant:
for elementidx = 1 : variables.Count
if strcmp(variables.values{elementidx}, 'Constant') %if you want to compare the value
%or maybe
if strcmp(variables.keys{elementidx}, 'Constant') %if you want to compare the actual key
  8 Comments
lotus whit
lotus whit on 3 Nov 2017
This is my all code :
filename = 'BA.xlsx';
% [num,txt,raw] = xlsread(filename)
[~, txt] = xlsread(filename,'Sheet1');
s_t= size(txt);
for ri=1:s_t(1)
for ci=1:s_t(2)
A=txt(ri,1);
B =txt(ri,2);
table_Ba{ri} = [A B];
end
end
% B1(:,1)= A;
B1=table_Ba{46};
B2= table_Ba{36};
B3=table_Ba{6};
B4=table_Ba{26};
keySet1 = {'1 0 1 0 1','1 1 1 1 1','1 0 0 0 1','1 0 0 0 0'};
valueSet1 = [B1,B2,B3,B4];
mapObj1 = containers.Map(keySet1,valueSet1)
Walter Roberson
Walter Roberson on 3 Nov 2017
valueSet1 = {B1,B2,B3,B4};
is my guess. If B1 and so on are character vectors, then [B1,B2,B3,B4] is a single concatenated character vector, which would not be appropriate for a list of keys like you have in keySet1.

Sign in to comment.

More Answers (1)

Walter Roberson
Walter Roberson on 30 Nov 2016
You create variables as an empty map. It has no keys. So your first for loop does nothing so no keys are created for variable
  1 Comment
jungdnb
jungdnb on 1 Dec 2016
Here is a screenshot from my matlab command line. First list are my variables.keys. Second list are the strings, in which I have to search for the variables.keys . Actually I've managed to modify the source code, so it does kind of the thing I want it to do. I want to test it for a bit, and will upload tomorrow, If I don't finde the right answer.

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!