Find MIN value among dates

[mark91345]

I have multiple "Tickers", but with different dates. I want to find the oldest date, based upon each ticker (Column C is the end result I am looking to get).
Note that sometimes, there is only one Ticker, while other times, there are 2, 3, or more. I cannot figure out what to do.


Office Excel 2016 (Windows) 32 bit


Excel 2016 (Windows) 32 bit
[Table="width:, class:head"][tr=bgcolor:#E0E0F0][th] [/th][th]

A​

[/th][th]

B​

[/th][th]

C​

[/th][/tr]
[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

1​

[/td][td]TICKER[/td][td]

DATE​

[/td][td]

Desired Result​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

2​

[/td][td]AA[/td][td]

04/18/2018​

[/td][td]

04/18/2018​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

3​

[/td][td]AAC[/td][td]

08/02/2017​

[/td][td]

08/02/2017​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

4​

[/td][td]AAC[/td][td]

11/01/2017​

[/td][td]

08/02/2017​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

5​

[/td][td]AAC[/td][td]

02/21/2018​

[/td][td]

08/02/2017​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

6​

[/td][td]AAC[/td][td]

05/02/2018​

[/td][td]

08/02/2017​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

7​

[/td][td]AAL[/td][td]

01/25/2018​

[/td][td]

01/25/2018​

[/td][/tr]


[tr=bgcolor:#FFFFFF][td=bgcolor:#E0E0F0]

8​

[/td][td]AAL[/td][td]

07/26/2018​

[/td][td]

01/25/2018​

[/td][/tr]
[/table][Table="width:, class:grid"][tr][td]Sheet: TEMP[/td][/tr][/table]

 

[sheetspread]

Excel 2010
	A	B	C
1	TICKER	DATE	Desired Result
2	AA	4/18/2018	4/18/2018
3	AAC	8/2/2017	8/2/2017
4	AAC	11/1/2017	8/2/2017
5	AAC	2/21/2018	8/2/2017
6	AAC	5/2/2018	8/2/2017
7	AAL	1/25/2018	1/25/2018
8	AAL	7/26/2018	1/25/2018
Sheet10

Cell Formulas
Range	Formula
C2	{=MIN(IF($A$2:$A$8=A2,$B$2:$B$8))}
Press CTRL+SHIFT+ENTER to enter array formulas.

 

[Aladin Akyurek]

If you have MINIFS on your system...

In C2 enter and copy down:

=MINIFS($B$2:$B$8,$A$2:$A$8,$A2)

 

[mark91345]

Thank you. It worked perfectly.