We assume that for $\sigma_c(Q)$...
+We assume that for $\sigma_c(Q)$ or $\delta_A(Q)$...
COUNT(*)
Estimating $\delta_A(Q)$ requires only COUNT(DISTINCT A)
Selectivity is a probability ($\texttt{SEL}(c, Q) = P(c)$)
Name | YearsEmployed | Role |
---|---|---|
'Alice' | 3 | 1 |
'Bob' | 2 | 2 |
'Carol' | 3 | 1 |
'Dave' | 1 | 3 |
'Eve' | 2 | 2 |
'Fred' | 2 | 3 |
'Gwen' | 4 | 1 |
'Harry' | 2 | 3 |
YearsEmployed | COUNT |
---|---|
1 | 1 |
2 | 4 |
3 | 2 |
4 | 1 |
COUNT(DISTINCT YearsEmployed) | $= 4$ |
MIN(YearsEmployed) | $= 1$ |
MAX(YearsEmplyed) | $= 4$ |
COUNT(*) YearsEmployed = 2 | $= 4$ |
YearsEmployed | COUNT |
---|---|
1-2 | 5 |
3-4 | 3 |
COUNT(DISTINCT YearsEmployed) | $= 4$ |
MIN(YearsEmployed) | $= 1$ |
MAX(YearsEmplyed) | $= 4$ |
COUNT(*) YearsEmployed = 2 | $= \frac{5}{2}$ |
YearsEmployed | COUNT |
---|---|
1-4 | 8 |
COUNT(DISTINCT YearsEmployed) | $= 4$ |
MIN(YearsEmployed) | $= 1$ |
MAX(YearsEmplyed) | $= 4$ |
COUNT(*) YearsEmployed = 2 | $= \frac{8}{4}$ |
Value | COUNT |
---|---|
1-10 | 20 |
11-20 | 0 |
21-30 | 15 |
31-40 | 30 |
41-50 | 22 |
51-60 | 63 |
61-70 | 10 |
71-80 | 10 |
SELECT … WHERE A = 33 |
+ $= \frac{1}{40-30}\cdot 30 = 3$ | +
SELECT … WHERE A > 33 |
+ $= \frac{40-33}{40-30}\cdot 30+22$ $\;\;\;+63+10+10$ $= 126$ | +