我希望使用R在具有相同长度的两个字符向量之间比较每个元素。我可以使用for循环轻松地做到这一点，但是我正在寻找一种替代方法。例如，
x<-c("A", "B", "C", "D", "E", "C")
y<-c("A", "B", "E", "D", "E", "C")
matches<-vector(length=6)
for(i in 1:6){
matches[i]<-match(x[i], y[i])
}
matches
[1] 1 1 NA 1 1 1
这似乎很简单，但是我很难弄清楚。
谢谢
答案 0 :(得分：1)
我们可以做library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use work.pkg.all;
entity porting is
-- Port ( );
end porting;
architecture Behavioral of porting is
component main is
port (
rstb: in std_logic;
clk: in std_logic;
en: in std_logic;
row: out integer;
col: out integer;
last_elem: out integer;
output_Data:out matrixData
);
end component;
component shift_reg is
port(
clk:in std_logic;
reset:in std_logic;
input:in integer;
q:out int_vector;
iter:out integer
);
end component;
signal data:matrixData;
signal row1: integer;
signal col1:integer;
signal element:integer;
signal clk:std_logic:='0';
signal enable:std_logic:='0';
signal reset:std_logic:='0';
signal input_elem:integer;
signal output_data:int_vector;
signal iteration: integer;
constant clk_period: time:=10 ns;
begin
p1: main port map('0','1',enable,row1,col1,element,data);
p2: shift_reg port map(clk,'0',input_elem,output_data,iteration);
clk_process :process
variable state:integer:=0;
variable r,c:integer;
begin
state:=state+1;
c:=state mod col1;
r:=state/col1;
input_elem<=data(r,c);
clk <= '0';
wait for clk_period/2;
clk <= '1';
wait for clk_period/2;
end process;
enable_process:process
begin
enable<='1';
wait for clk_period*52538;
enable<='0';
end process;
end Behavioral;
来逐元素比较两个向量
==对于匹配值，它返回x == y
，对于不匹配，它返回TRUE
如果我们需要转换为FALSE和1
NA