data tabela_a;
    do i=1 to 1000;
    kolumna_a=1;
    output;
    end;
run;

data tabela_b;
    do i=1 to 1500 by 10;
    kolumna_b='AAA';
    output;
    end;
run;

*³ıczenie zbiorów;
data z1;
    set tabela_a tabela_b indsname=zrodlo;
    zbior=zrodlo;
run;

data z1;
    set tabela_a(in=in1) tabela_b(in=in2);
    z1=in1;
    z2=in2;
run;

*interleaving;
proc sort data=tabela_a out=tabela_a;
    by i;
run;
proc sort data=tabela_b out=tabela_b;
    by i;
run;
data z2;
    set tabela_a tabela_b;
    by i;
run;

*merge;
*dane musza byc posortowane;
proc sort data=tabela_a out=tabela_a;
    by i;
run;
proc sort data=tabela_b out=tabela_b;
    by i;
run;

data z3;
    merge tabela_a tabela_b;
run;

*merge + klucz;
*dane musza byc posortowane;
*identyczne nazwy kolumn do laczenia;
data z4;
    merge tabela_a(in=in1) tabela_b(in=in2);
    by i; *klucz;
    z1=in1;
    z2=in2;
run;

*inner join;
data z4;
    merge tabela_a(in=in1) tabela_b(in=in2);
    by i; *klucz;
    if in1=1 and in2=1;
run;

*left join;
data z4;
    merge tabela_a(in=in1) tabela_b(in=in2);
    by i; *klucz;
    if in1=1;
run;

*klucze, ktore sie nie lacza;
data z6;
    merge tabela_a(in=in1) tabela_b(in=in2);
    by i; *klucz;
    if in1=0 or in2=0;
run;

*klucze wystepujace w tabela_a, ale nie wystapily w tabela_b;
data z7;
    merge tabela_a(in=in1) tabela_b(in=in2);
    by i; *klucz;
    if in1=1 and in2=0;
run;

data wynik;
    set tabela_a;
run;

proc append base=wynik data=tabela_b force;
run;

libname X 'C:\Users\student\Desktop\zbiory_laczenie_sas';
data zz1;
    set X.popularnames_2011 X.popularnames_2012 X.popularnames_2013 X.popularnames_2014 indsname=in;
    rok=in;
run;