PATH=?‘e:\xindian\data3‘;?%?path?where?data?are?saved
HEADERFILE=?‘116.hea‘;??????%?header-file?in?text?format
ATRFILE=?‘116.atr‘;?????????%?attributes-file?in?binary?format
DATAFILE=‘116.dat‘;?????????%?data-file
SAMPLES2READ=650000;?????????%?number?of?samples?to?be?read
????????????????????????????%?in?case?of?more?than?one?signal:
????????????????????????????%?2*SAMPLES2READ?samples?are?read

%------?LOAD?HEADER?DATA?----------
fprintf（1‘\\n$>?WORKING?ON?%s?...\n‘?HEADERFILE）;
signalh=?fullfile（PATH?HEADERFILE）;
fid1=fopen（signalh‘r‘）;
z=?fgetl（fid1）;
A=?sscanf（z?‘%*s?%d?%d?%d‘[13]）;
nosig=?A（1）;??%?number?of?signals
sfreq=A（2）;???%?sample?rate?of?data
clear?A;
for?k=1:nosig
????z=?fgetl（fid1）;
????A=[];
????A=?sscanf（z?‘%*s?%d?%d?%d?%d?%d‘[15]）;
????dformat（k）=?A（1）;???????????%?format;?here?only?212?is?allowed
????gain（k）=?A（2）;??????????????%?number?of?integers?per?mV
????bitres（k）=?A（3）;????????????%?bitresolution
????zerovalue（k）=?A（4）;?????????%?integer?value?of?ECG?zero?point
????firstvalue（k）=?A（5）;????????%?first?integer?value?of?signal?（to?test?for?errors）
end;
fclose（fid1）;
clear?A;

%------?LOAD?BINARY?DATA?---------
if?dformat~=?[212212]?error（‘this?script?does?not?apply?binary?formats?different?to?212.‘）;?end;
signald=?fullfile（PATH?DATAFILE）;????????????%?data?in?format?212
fid2=fopen（signald‘r‘）;
A=?fread（fid2?[3?SAMPLES2READ]?‘uint8‘）‘;??%?matrix?with?3?rows?each?8?bits?long?=?2*12bit?A?是三列3000行的矩陣
fclose（fid2）;
M2H=?bitshift（A（:2）?-4）;????????%字節(jié)向右移四位，取每一行中的第二列的字節(jié)的高四位，即取字節(jié)的高四位，比如56（112000），右移四位就是（00000011）為3
M1H=?bitand（A（:2）?15）;??????????%取已經(jīng)移動(dòng)了的字節(jié)的低四位，就是原來字節(jié)的低四位，bit（ab）ab的二進(jìn)制對應(yīng)位求與運(yùn)算，比如bitand（3（0011）10（2000））之后為（0010），即為2
PRL=bitshift（bitand（A（:2）8）9）;?????%?sign-bit???取出字節(jié)低四位中最高位，向左移九位
PRR=bitshift（bitand（A（:2）128）5）;???%?sign-bit???取出字節(jié)高四位中最高位，向左移五位
M（?:??1）=?bitshift（M1H8）+?A（:1）-PRL;
M（?:??2）=?bitshift（M2H8）+?A（:3）-PRR;
if?M（1:）?~=?firstvalue?error（‘inconsistency?in?the?first?bit?values‘）;?end;%判斷與第一個(gè)值是否一樣
switch?nosig??%看是不是兩個(gè)采樣信號(hào)
????
case?2
????M（?:??1）=?（M（?:??1）-?zerovalue（1））/gain（1）;
????M（?:??2）=?（M（?:??2）-?zerovalue（2））/gain（2）;
????TIME=（0:（SAMPLES2READ-1））/sfreq;%sfreq是采樣率，除以采樣率
case?1
????M（?:??1）=?（M（?:??1）-?zerovalue（1））;
????M（?:??2）=?（M（?:??2）-?zerovalue（1））;
????M=M‘;
????M（1）=[];
????sM=size（M）;
????sM=sM（2）+1;
????M（sM）=0;
????M=M‘;
????M=M/gain（1）;
????TIME=（0:2*（SAMPLES2READ）-1）/sfreq;%time是3000個(gè)采樣點(diǎn)的時(shí)間位置
otherwise??%?this?case?did?not?appear?up?to?now!
????%?here?M?has?to?be?sorted!!!
????disp（‘Sorting?algorithm?for?more?than?2?signals?not?programmed?yet!‘）;
end;
clear?A?M1H?M2H?PRR?PRL;
fprintf（1‘\\n$>?LOADING?DATA?FINISHED?\n‘）;

%------?LOAD?ATTRIBUTES?DATA?---------
atrd=?fullfile（PATH?ATRFILE）;??????%?attribute?file?with?annotation?data
fid3=fopen（atrd‘r‘）;
A=?fread（fid3?[2?inf]?‘uint8‘）‘;%A有2121行，2列
fclose（fid3）;
ATRTIME=[];?%保存每一個(gè)心跳標(biāo)記的時(shí)間點(diǎn)
ANNOT=[];%保存每個(gè)心跳的注釋代碼
sa

?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----

?????文件???????4759??2014-04-28?19:25??Untitled.m

-----------?---------??----------?-----??----

?????????????????4759????????????????????1