Switch Individual Lights In LED Series
Placing a switch ACROSS lamp1 will extinguish it when the switch is closed.
The current in the "leg" will increase.
You can design the cct so current is "somewhat low when switch is open and somewhat high when switch is closed so brightness is "much the same" overall. 

eg with 3V LEDs.  If series R is 100 Ohms. 


2 LEDS I = V/R = (12-9)/100 = 30 mA. 
2 LEDS I = V/R = (12-6)/100 = 60 mA. 


You'll notice a 2:1 difference but it will not APPEAR to be that much. 

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Alternative - 

Use a SPDT switch and operate EITHER the LED or a zener of equal voltage.
LEDS shown are example only.  

With eg 3V white LEDs I ~~= V/R = (12 - 9) /100 = 30 mA.
Choose zener to suit LED.
Zener voltage will depend on current.
eg for a 2V7 rated zener, Vz is usually about 2V7 at 10 mA and somewhat more at higher currents.

As jsotola notes -  a resistor could be used in place of the zener.
R = V/I  = V_LED_replaced / Istring 
For white LEDs at Vf = 3V and 30 mA LED string current,
R = V/I = 3V / 30 mA = 100 ohms
So 3 LEDs have 100 ohms in series and 2 LEDS have 100 ohms in series.
(Following that line of though 4 LEDS need NO Ohms in series at 12V BUT this is "not really so". The equation relies on Vsupply being exactly 12V and each LED having a Vf (forward voltage) of 3V at 30 mA. If that was exactly and always so then it would work. In practice the spread in LED Vfs and small supply variations would usually lead to "very poor" results. 







simulate this circuit &ndash; Schematic created using CircuitLab