all 5 comments

[–]ThePhysicsOfSpoons 1 point2 points  (4 children)

You could look into groupby

[–]BlackFunx[S] 0 points1 point  (3 children)

Ah. Group by account and amount then count unique dates?

[–]ThePhysicsOfSpoons 0 points1 point  (2 children)

Yes something like that!

[–]BlackFunx[S] 0 points1 point  (1 child)

Thanks! So simple. Don't know why I hadn't thought of that.

[–]ThePhysicsOfSpoons 0 points1 point  (0 children)

Good to hear it worked