A good book for new python learners

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.

怎么在免安装版本的sas当中读入sas7bdat文件

如果是安装版的sas，不管是9.1还是9.2，都能双击打开sas7bdat文件。但是免安装的sas 9.1.3 直接打开是不行的，通过import什么的也不可以。可以这么做：

比如把 temp.sas7bdat 位置在 "c:\mydata\temp.sas7bdat" ，那么：

在sas中输入  libname mylib 'c:\mydata';

然后再sas右侧，explore，libraries， 看见mydata，然后点击进去，就看见自己的data了。

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;

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;

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;