Forum Discussion
graphic representation
| ID | test_A | test_max | test_M |
| 125 | 0 | 0 | 0 |
| 125 | 10000 | 15000 | 12 |
| 125 | 20000 | 25000 | 24 |
| 125 | 30000 | 35000 | 36 |
| 140 | 0 | 0 | 0 |
| 140 | 14000 | 10000 | 12 |
| 140 | 20000 | 25000 | 24 |
| 870 | 0 | 0 | 0 |
| 870 | 10000 | 15000 | 12 |
| 870 | 12000 | 25000 | 24 |
| 870 | 14000 | 30000 | 36 |
| 870 | 15000 | 32000 | 48 |
| 900 | 0 | 0 | 0 |
| 900 | 12000 | 24000 | 12 |
| 900 | 13000 | 22000 | 24 |
| 900 | 14000 | 35000 | 36 |
a)
draw the curvesb) tracer la ligne droite qui part de 0 et va jusqu'à la valeur maximale de test_max
ex : si dans mon filtre id je selectionne un ID je dois voir la corbe de test_A et la droite qui part de 0 jusqu'a la valeur maximale de Test_max
c) if I select two different IDs which do not have the same identical line, a message is displayed "not compatible", in this case the line is not displayed and only the curves which are displayed
ex: ID 125 and ID 900 have the same maximum value 35000 so I will see the curve and the straight line, on the other hand ID 125 and 140 do not have the same maximum value, in this case I select these two, I would not have the right and a message "not compatible is displayed" and only the curves will also be displayed
d)put a color if my curve is above the line it's red, if it's below it's green
THanks
Super Userd) assumes that the latest data point is the highest. That assumption fails for ID 900 and Month 24, for example.
Might be better to just add a trend line.
for c) I would use small multiples
lbendlin thank you for your feedback, the problem is that my trend line must pass through the origin (0,0), for each ID up to its maximum value of test_max, and it's on a single chart, because I have multiple ids