how add "$" and "' ' " in array string

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piero
piero 2023-8-16
>> T(:,2)
ans =
19×1 string array
"139411.39"
"115944.39"
"413970.912"
"124256.379"
"144673.585"
"93473.162"
"334232.706"
"105488.574"
"114121.302"
"126438.346"
"-11956.632"
"95737.662"
"169120.64"
"-43385.61"
"215426.368"
"-137202.827"
"70333.129"
"-71453.588"
"47995.706"
i want display this uitable app designer (i want $ and ' for separator 000)
"$ 139'411.39"
"$ 115'944.39"
  2 个评论
Voss
Voss 2023-8-16
For negative values, do you want the minus sign before or after the dollar sign? That is, should it be "-$43385.61" or "$-43385.61"?
piero
piero 2023-8-16
编辑:piero 2023-8-16
"$-43385.61" thank
T originally is array double

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回答(3 个)

Dyuman Joshi
Dyuman Joshi 2023-8-16
编辑:Dyuman Joshi 2023-8-16
T = ["139411.39"
"115944.39"
"413970.912"
"124256.379"
"144673.585"
"93473.162"
"334232.706"
"105488.574"
"114121.302"
"126438.346"
"-11956.632"
"95737.662"
"169120.64"
"-43385.61"
"215426.368"
"-137202.827"
"70333.129"
"-71453.588"
"47995.706"];
%Split into integer part and decimal part
T = split(T,".");
%Add apostrophe for thousand's place in the integer part
%as mentioned in the problem statement
T(:,1) = regexprep(T(:,1),'(\d+)(\d{3})$',"$1'$2");
%Join the table by columns
%and add the Dollar sign
T = "$ " + join(T,".",2)
T = 19×1 string array
"$ 139'411.39" "$ 115'944.39" "$ 413'970.912" "$ 124'256.379" "$ 144'673.585" "$ 93'473.162" "$ 334'232.706" "$ 105'488.574" "$ 114'121.302" "$ 126'438.346" "$ -11'956.632" "$ 95'737.662" "$ 169'120.64" "$ -43'385.61" "$ 215'426.368" "$ -137'202.827" "$ 70'333.129" "$ -71'453.588" "$ 47'995.706"
  10 个评论
Stephen23
Stephen23 2023-8-17
Note that this code includes at most one apostrophe, i.e. it misses the apostrophes for values >=1e6:
prof = 9876543210;
T1 = string(fix(prof));
T2 = erase(compose("%g",mod(prof,1)),"0.");
%Add separator for thousand place
T1 = regexprep(T1,'(\d+)(\d{3})$',"$1'$2");
%join the strings
out = "$ "+ T1 + "." + T2
out = "$ 9876543'210.0"
Dyuman Joshi
Dyuman Joshi 2023-8-17
@Stephen23, Yes I am aware of that, as that is what OP has stated in the problem above and I have mentioned that in the comment as well.
In case OP wants to have separator for every thousand's place, this would be a simple approach -
prof = [9873210.123 123456 -randi([1e8 1e9])/1e3 -0.2357 6.66 0.42069]';
T1 = fix(prof);
T2 = erase(string(abs(prof - T1)),"0.");
out = "$ "+arrayfun(@ThousandSep, T1)+"."+T2
out = 6×1 string array
"$ 9'873'210.123" "$ 123'456.0" "$ -415'537.916" "$ -0.2357" "$ 6.66" "$ 0.42069"
function out = ThousandSep(in)
%THOUSANDSEP adds thousands Separators to a 1x1 array.
import java.text.*
v = DecimalFormat;
out = replace(string(v.format(in)),",", "'");
end

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Stephen23
Stephen23 2023-8-17
编辑:Stephen23 2023-8-17
format long G
V = [0;-23;123.456;-0.78;9;1234567.89;-987654321;7;-54321]
V = 9×1
1.0e+00 * 0 -23 123.456 -0.78 9 1234567.89 -987654321 7 -54321
S = compose("$ %.2f",V(:));
S = regexprep(S,"(\d{1,3})(?=(\d{3})+(\D|$))","$1'")
S = 9×1 string array
"$ 0.00" "$ -23.00" "$ 123.46" "$ -0.78" "$ 9.00" "$ 1'234'567.89" "$ -987'654'321.00" "$ 7.00" "$ -54'321.00"

Mrutyunjaya Hiremath
% Sample data
T = [
"139411.39";
"115944.39";
"413970.912";
"124256.379";
"144673.585";
"93473.162";
"334232.706";
"105488.574";
"114121.302";
"126438.346";
"-11956.632";
"95737.662";
"169120.64";
"-43385.61";
"215426.368";
"-137202.827";
"70333.129";
"-71453.588";
"47995.706";
];
% Convert to numerical array
T_num = str2double(T);
% Format the numbers with custom formatting
formattedData = strings(size(T));
for i = 1:length(T_num)
if T_num(i) >= 0
formattedData(i) = sprintf("$ %d,%03d.%02d", floor(T_num(i)/1000), mod(floor(T_num(i)),1000), round(100*mod(T_num(i),1)));
else
T_abs = abs(T_num(i));
formattedData(i) = sprintf("-$ %d,%03d.%02d", floor(T_abs/1000), mod(floor(T_abs),1000), round(100*mod(T_abs,1)));
end
end
disp(formattedData)
"$ 139,411.39" "$ 115,944.39" "$ 413,970.91" "$ 124,256.38" "$ 144,673.58" "$ 93,473.16" "$ 334,232.71" "$ 105,488.57" "$ 114,121.30" "$ 126,438.35" "-$ 11,956.63" "$ 95,737.66" "$ 169,120.64" "-$ 43,385.61" "$ 215,426.37" "-$ 137,202.83" "$ 70,333.13" "-$ 71,453.59" "$ 47,995.71"
% Suppose 'app' is your app struct and 'UITable' is the name of your uitable component
% You can set the Data property as follows:
app.UITable.Data = formattedData;
  1 个评论
piero
piero 2023-8-16
编辑:piero 2023-8-16
I don't think it's the most efficient solution
there is the word reserved "compose" but I don't know how to use it

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