Survival Calculations I am trying to calculate survival rate for all grades (Kindergarten to 1st Grade) of school. See the example below for Grade 1 students. Y0 refers to number of students enrolled in grade 1 in that particular year. Y1 : how many of them are still in the school 1 year later. These students would have moved to grade 2 in Y1. Similarly for the other years. I have similar data for other grades as well for years 2010 to 2019. As of now I am calculating survival rate based on (number of students in Yn divided by number of students in Y0). For example : In Y2 survival rate is column sum of Y2 divided by column sum of Y0 (excluding 2018 as data is not available in Y2 for 2018). It is 76%. I will do the same calculations for other Yn periods and for other grades and then average out all the annual rates.
I wanted to know is there any better way to calculate survival rate more statistically?
 A: The Kaplan–Meier estimator for survival probability, $S[t]$, is computed as a product of survival probabilities over disjoint sub-intervals
$$
S[t] = \prod_{k \le t} s_k
$$
where $s_k$ is the probability to survive over $t=[k-1,k)$.
In your case, the atomic sub-intervals correspond to columns of your table $n_{:,k}$.
For a single row (i.e. cohort), $i$, you could estimate the interval probabilities as
$$
s_k^{(i)} = \frac{n_{i,k}}{n_{i,k-1}}
$$
and in this case, the Kaplan-Meier estimate
$$
S_k^{(i)} = \frac{n_{i,1}}{n_{i,0}}\times\frac{n_{i,2}}{n_{i,1}}\times\cdots\times\frac{n_{i,k-1}}{n_{i,k-2}}\times\frac{n_{i,k}}{n_{i,k-1}} = \frac{n_{i,k}}{n_{i,0}}
$$
reduces to a simple ratio, due to cancellation of successive factors.
However to use all of the data, here you want to pool or aggregate the different cohorts (rows). As noted in the comments, you cannot simply sum each column, because for later cohorts the data is censored.
For computing aggregate interval probabilities, both numerator and denominator should be summed over the same "lumped cohort" (set of rows), corresponding to where there is data for both columns.
Note that once you do this, the product of the interval probabilities will no longer reduce to a simple ratio (i.e. the denominator for one interval will no longer cancel with the following numerator).
The need for the multiplicative approach is illustrated below (using your table):

The blue curve uses the Kaplan–Meier approach, while the orange curve uses a single ratio of current year to Y0 (both summed over rows with current-year data). Note that while the two are similar, the naive ratio approach shows a logically impossible increase in survival from Y8 to Y9. In contrast, the Kaplan–Meier estimate correctly shows constant survival from Y6 to Y9 (i.e. since  after Y6, there is no decrease in survivors in any row).

Example Calculations
(Note: survival at Y0 is $s_0 = 1$, by definition.)
Survival from Y0 to Y1 ...
$$
s_1 = \frac{\sum_{2010\le{i}\le2018}n_{i,1}}{\sum_{2010\le{i}\le2018}n_{i,0}} = \frac{3+\cdots+114}{3+\cdots+98} = \frac{599}{686}
$$
Survival from Y1 to Y2 ...
$$
s_2 = \frac{\sum_{2010\le{i}\le2017}n_{i,2}}{\sum_{2010\le{i}\le2017}n_{i,1}} = \frac{3+\cdots+104}{3+\cdots+114} = \frac{435}{501}
$$
...
Survival from Y5 to Y6 ...
$$
s_6 = \frac{\sum_{2010\le{i}\le2013}n_{i,6}}{\sum_{2010\le{i}\le2013}n_{i,5}} = \frac{2+\cdots+28}{2+\cdots+31} = \frac{59}{68}
$$
...
Survival from Y8 to Y9 ...
$$
s_9 = \frac{\sum_{2010\le{i}\le2010}n_{i,9}}{\sum_{2010\le{i}\le2010}n_{i,8}} = \frac{2}{2}
$$
(Note: $s_7 = s_8 = s_9 = 1$, since there is no rightward decrease in students from Y6 onward.)
Similarly, $S_0 = s_0$, $S_1 = s_0s_1$, $S_2 = s_0s_1s_2$, etc.
For example
$$
S_2 = 1 \times \frac{599}{686} \times \frac{435}{501} \approx 0.758
$$
