PHP:检测2个多维数组中的唯一值和重复值

So I searched around, but I couldn't find a good answer to my question, so here goes. I'm pulling some data through json, which gets updated every hour.

I want to take this data and save it to my database and categorize the data as "new", "active" & "old"

I have saved a json feed in my database that looks like this when I pull it from the database:

Array ( 
[0] => Array ( [0] => Blood [1] => all [2] => 2040154277 [3] => 74708 [4] => a )
[1] => Array ( [0] => Blood [1] => all [2] => 2040088680 [3] => 42973 [4] => b ) 
[2] => Array ( [0] => Blood [1] => all [2] => 2038497801 [3] => 21308 [4] => c ) 
[3] => Array ( [0] => Blood [1] => all [2] => 2040404080 [3] => 82800 [4] => d ) 
[4] => Array ( [0] => Blood [1] => all [2] => 2039988402 [3] => 74703 [4] => e ) 
[5] => Array ( [0] => Blood [1] => all [2] => 2038753364 [3] => 82800 [4] => f ) 
[6] => Array ( [0] => Blood [1] => all [2] => 2040065624 [3] => 54443 [4] => g ) 
[7] => Array ( [0] => Blood [1] => all [2] => 2040153539 [3] => 82800 [4] => h )
)

I have made it a little bit more readable by adding letters to [4], but the only unique identifier is the number in [2].

Anyway, this data reflects what data was current 1h ago, so when the new feed comes in:

Array ( 
[0] => Array ( [0] => Blood [1] => all [2] => 2040123457 [3] => 74708 [4] => aa )
[0] => Array ( [0] => Blood [1] => all [2] => 2040154277 [3] => 74708 [4] => a )
[1] => Array ( [0] => Blood [1] => all [2] => 2040088680 [3] => 42973 [4] => b ) 
[3] => Array ( [0] => Blood [1] => all [2] => 2040404080 [3] => 82800 [4] => d ) 
[4] => Array ( [0] => Blood [1] => all [2] => 2039988402 [3] => 74703 [4] => e ) 
[5] => Array ( [0] => Blood [1] => all [2] => 2038753364 [3] => 82800 [4] => f ) 
)

We notice that [4] c, g & h have expired, but also that [4] aa is new.

Using the old & new array, how can I identify which ones are new, active & old for a simple insert/update in my sql in a fairly efficient way as the multidimensional array can very well contain 50-60k arrays.

Is there a good way to solve this?