Merging standard deviation of two groups with duplicates I am using Parallel Algorithm (http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Parallel_algorithm) to merge standard deviation of two groups 
Can we merge two groups if there are duplicate entries in them
example I have two groups group1 and group2 having userName and usage
group1
 user1 10
 user2 12
 user3 15
group2
 user2 14
 user3 13
 user4 16
merged
 user1 10
 user2 26
 user3 28
 user4 16
stdev - 7.3485
i want to find st dev of merged group, but problem is because of high data volume I can't maintain userName , so i am maintaining sum and user count, and by probablistic counting algorithm I can find unique users also
group1
 count-3, sum-37, M2-9.935
group2
 count-3, sum-43, M2-7.919
merged group
 unique- 4
 group1:: count-3, sum-37, M2-9.935
 group2:: count-3, sum-43, M2-7.919
now with this information available, can i find standard deviation of merged group??
 A: You can't calculate what you need from the counts and variances of the two groups. Consider the following example.
On day $1$ your data is this:
$$\begin{array}{rl|rl} \text{Group }1 & & \text{Group } 2 & \\
\text{User }1 & 1  & \text{User }1& 9  \\
\text{User }2 & 10  & \text{User }3& 10  \\
\text{User }4 & 9  & \text{User }4& 1  \\
\end{array}$$
So both groups have mean $\frac{20}3$ and standard deviation $\sqrt{\frac{1238}9}$. 
On  day $2$ the numbers are the same but the users are different
$$\begin{array}{rl|rl} \text{Group }1 & & \text{Group } 2 & \\
\text{User }1 & 10  & \text{User }1& 10  \\
\text{User }2 & 1  & \text{User }3& 1  \\
\text{User }4 & 9  & \text{User }4& 9  \\
\end{array}$$
So the means and standard deviations haven't changed for the individual groups.
The list of unique users is the same for both days too. But look what happens when I combine the groups.
$$\begin{array}{rl|rl} \text{Day }1 & & \text{Day } 2 & \\
\text{User }1 & 10  & \text{User }1& 20  \\
\text{User }2 & 10  & \text{User }2& 1  \\
\text{User }3 & 10  & \text{User }3& 1  \\
\text{User }4 & 10  & \text{User }4& 18  \\
\end{array}$$
The groups have the same mean, but the standard deviations are wildly different. 
I don't know what you can do from here.  The numbers you've got aren't enough to tell you the standard deviation. Your algorithm has to be able to distinguish between the two examples above, so you need to keep track of the user name. 
I think you're going to have to merge the two data sets. 
