R: get one data frame from the zoo object list

The input displayed at the top of the question is represented by the first three input components listed at the bottom of the question. The variable name used at the bottom of the question lodf

seems to suggest it contains a list of data frames, but it actually contains a list of zoo objects.

The question asks one result with a dataframe, but we assume the output should also be one zoo series for consistency. We will also use the name L

for input as it will lodf

incorrectly suggest a list of data frames. If z

is the result in the form of a zoo row, then

data.frame(index = index(z), data = coredata(z))

      

        
        
        
      

    

can be used if a data frame is really needed.

In the output section at the end of this answer, we will show the result of using as our input L <- lodf[1:3]

(i.e. only the first 3 components) and separately show the output using L <- lodf

(i.e. all components) as our input.

1) Reduce . We concatenate a series of zoos in a list, L

return the list and fill in the missing values ​​with 0. Then use Reduce

to sum the components:

Reduce(`+`, do.call(merge, c(L, retclass = "list", fill = 0)))

      

        
        
        
      

    

1a) . A variation on this is to return the zoo object from merge

(which is the default if we don't specify retclass

), then fill it with NA with 0, turn it back into a list, and use Reduce

:

Reduce(`+`, as.list(na.fill(do.call(merge, L), 0)))

      

        
        
        
      

    

2) rowSums . In this solution, we concatenate lists to provide a zoo object z

, optionally add the column names, and then add rows that produce the zoo target.

z <- do.call(merge, L)
colnames(L) <- seq_along(L) # optionally add names
zoo(rowSums(z, na.rm = TRUE), time(z))

      

        
        
        
      

    

Please note that previously a rowSums

solution to the zoo objects was found here

3) + If we knew that there were exactly 3 components in the list, then an alternative way to write above would be this. We optionally add names 1, 2, 3, concatenate the zoo objects and fill NA with 0. Finally, we add the series together. Modify in an obvious way if the number of components is different.

z0 <- na.fill(do.call(merge, L), 0)
colnames(z0) <- 1:3 # optionally add names 1, 2, 3
z0[, 1] + z0[, 2] + z0[, 3]

      

        
        
        
      

    

Output Using L <- lodf[1:3]

as shown at the beginning of the question, where lodf

displayed at the bottom of the question, our output is:

2012-11-01 2012-11-02 2013-07-12 2013-08-13 2013-09-11 2014-07-01 2014-07-22 
      1.00       1.27       0.91       1.00       0.99       0.98       0.50 
2014-07-24 2014-08-08 2014-08-14 2014-08-20 2014-08-27 2014-09-12 
      0.75       1.15       2.61       1.36       1.28       3.67 

      

        
        
        
      

    

or using L <- locf

the above, we get the following (except for solution 3, which would have to be modified in an obvious way to use 6 rather than 3 components):

2012-11-01 2012-11-02 2012-11-06 2012-11-07 2012-11-08 2012-11-16 2012-11-21 
      7.52       1.27       3.44       1.29       1.35       0.66       1.28 
2012-12-05 2012-12-12 2012-12-20 2013-02-04 2013-02-05 2013-02-06 2013-02-07 
      1.17       0.59       3.41       4.30       3.11       2.55       2.65 
2013-02-11 2013-02-13 2013-02-14 2013-02-15 2013-02-19 2013-02-20 2013-02-22 
      0.00       2.87       3.14       3.24       3.41       4.04       4.19 
2013-02-26 2013-03-20 2013-04-02 2013-04-05 2013-04-10 2013-04-15 2013-04-29 
      4.34       4.44       0.66       1.20       1.30       1.29       1.30 
2013-05-10 2013-05-13 2013-05-15 2013-05-16 2013-05-22 2013-05-24 2013-06-04 
      1.27       0.77       0.69       1.16       0.55       0.00       1.26 
2013-06-11 2013-06-20 2013-07-01 2013-07-04 2013-07-11 2013-07-12 2013-07-15 
      1.23       1.28       1.23       1.17       0.66       0.91       1.18 
2013-07-19 2013-07-22 2013-07-23 2013-07-24 2013-07-25 2013-07-29 2013-08-09 
      0.66       1.29       1.35       1.45       1.53       1.61       1.82 
2013-08-13 2013-08-14 2013-08-15 2013-08-21 2013-09-05 2013-09-06 2013-09-11 
      1.00       1.80       1.89       1.80       1.81       1.78       0.99 
2013-09-27 2013-09-30 2013-10-01 2013-10-03 2013-10-04 2013-10-07 2013-10-08 
      1.68       2.18       1.68       1.56       1.93       1.84       1.69 
2013-10-09 2013-10-23 2013-10-25 2013-10-28 2013-10-30 2013-11-01 2013-11-05 
      1.18       1.73       1.18       1.72       1.83       1.90       1.99 
2013-11-07 2013-11-12 2013-11-13 2013-11-14 2013-11-18 2013-11-20 2013-11-22 
      1.93       1.87       1.96       2.10       2.22       2.33       2.38 
2013-12-06 2013-12-13 2013-12-20 2014-01-02 2014-01-07 2014-01-08 2014-01-14 
      2.35       2.23       2.16       2.18       2.17       2.20       2.29 
2014-01-21 2014-01-28 2014-01-30 2014-01-31 2014-02-05 2014-03-03 2014-03-04 
      2.27       2.28       2.42       2.48       2.99       2.56       2.65 
2014-03-05 2014-03-10 2014-03-17 2014-03-21 2014-03-24 2014-03-28 2014-03-31 
      2.69       3.21       2.70       2.80       2.79       2.80       2.78 
2014-04-03 2014-04-07 2014-04-09 2014-04-10 2014-04-11 2014-04-15 2014-04-17 
      2.26       2.78       2.26       2.12       2.07       1.97       1.84 
2014-04-22 2014-04-24 2014-05-22 2014-05-26 2014-05-30 2014-06-03 2014-06-27 
      1.77       1.18       1.70       1.78       1.91       1.98       1.93 
2014-07-01 2014-07-03 2014-07-07 2014-07-10 2014-07-22 2014-07-24 2014-08-08 
      0.98       1.83       1.76       1.18       0.50       0.75       1.15 
2014-08-14 2014-08-15 2014-08-19 2014-08-20 2014-08-27 2014-08-28 2014-09-04 
      2.61       1.01       0.97       1.36       2.14       0.69       0.56 
2014-09-12 

      

        
        
        
      

    

Updates . Additional solutions have been added and presentations reorganized and expanded.