# Head First Programming: A Learner's Guide to Programming Using the Python Language

### Find the max length of non-repeated substring

The question is: given the string such as 'abcabcbb', 'bbbbb', 'abbcbba', what is the max length of non-repeated sub-string? For example, for 'abcabcbb', the answer is the length of 'abc' or 'bca' or 'cab', they all have length 3; for 'bbbbb', it only has length 1; for 'abbcbba', its length is 2.

## Thursday, January 27, 2011

### zz: Change the variable value for consecutive variable with some special conditions

This question is answered by other people not me.

The data looks like below. Want to do: for the variable high, if there are consecutive 1s or consecutive -1s, we will compare the corresponding p_sm. For 1, keep the maximum p_sm and set the others as missing . ; for -1, keep the minimum and set the others as missing .;  For example, for the first and second obs with consecutive 1, since 102.88<105.29, we will change the second 1 as .;  In other words, We need to change 2, 14, 16, 20 and so on to be missing.

There are two methods given by elek:

* method 1;
data want;
retain min max;
do until (last.high);
set tmp1.have;
by high notsorted;
if first.high then do;
min=p_sm;
max=p_sm;
end;
else do;
min=min(min,p_sm);
max=max(max,p_sm);
end;
end;
do until (last.high);
set tmp1.have;
by high notsorted;
if high=1 and p_sm^=max then high=.;
if high=-1 and p_sm^=min then high=.;
output;
end;
run;

* method 2;
data tmp1;
set tmp1.have;
obs=_n_;
run;
proc means data=tmp1 noprint idmin;
by high notsorted;
var p_sm;
id  p_sm obs;
output out=tmp2(drop=_freq_ _type_) min(p_sm)=min max(p_sm)=max;
run;
proc means data=tmp1 noprint;
by high notsorted;
var p_sm;
id  p_sm obs;
output out=tmp3(drop=_freq_ _type_) min(p_sm)=min max(p_sm)=max;
run;
data want;
merge tmp1 tmp2 tmp3;
by obs;
if high=1 and max^=p_sm then high=.;
if high=-1 and min^=p_sm then high=.;
drop min max obs;
run;

## Wednesday, January 26, 2011

### If the difference of two consecutive number is less then 7, then assign it as missing

options formdlim=' ';

/***************************************************************************
If the difference of two consecutive number is less then 7, then assign it as missing;
*  First, the target numbers will be set as 0; If set as missing ,there is problem;
*  Next, set the zeros to be missing.
***************************************************************************/

data temp;
input d1-d6;
cards;
1 2 7 23 100 1000
2 3 33 54 56 1000
3 . 4 6 44 100
;
run;

data temp1;
set temp;
array ss(1:6) d1-d6;
array s(1:6) d1-d6;
do i=2 to 6;
if s[i]-s[i-1]<=7 then do;
ss[i]=0/* Here if set as missing, there is problems */
ss[i-1]=0;
end;
end;
run;

data want;
set temp1;
array s d1-d6;
do i=1 to 6;
if s[i]=0 then s[i]=.;
end;
run;

proc print data=want;
run;

## Saturday, January 22, 2011

### Get the different chars appearance frequency

options formdlim='-';

************* Another Problem form MITBBS *****************;
* The question like this: have data x like below consists of 1 and 2 only *;
* Want to count the frequency of 1's and 2's appear consecutively *;
* More detailed, how many times 1 appear once consecutively *;
* More detailed, how many times 1 appear twice consecutively *;
* And so on *;
* Until how many times 2 appear ** times consecutively ? *;
* That is, the result should be: *;
/**************************************************************
x   counter      freq
----------------------------
1         1         3
1         4         1
2         2         2
2         3         1
2         5         1
**************************************************************/

data test;
input x;
cards;
1
2
2
1
1
1
1
2
2
2
2
2
1
2
2
1
2
2
2
;
run;

data new;
set test;
by x notsorted;
retain counter;
if first.x then counter=0;
counter+1;
if last.x then output;
run;

proc print data=new;
run;

proc sql;
select x, counter, n(counter) as freq
from new
group by x, counter
order by x, counter
;
quit;

***  If don't use proc sql, we can use these data steps: **;
data new;
set test;
by x notsorted;
retain counter;
if first.x then counter=0;
counter+1;
if last.x then output;
run;

proc print data=new;
run;

proc sort data=new;
by x  counter;
run;

data want;
set new;
by x  counter;
retain freq;
if first.x  or first.counter then  freq=1;
else freq=freq+1;
if last.x or last.counter then output;
run;

proc print data=want;
run;