在R中更改数据帧

Question

I have a data frame that has the first column go from 1 to 365 like this

c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2...

and the second column has times that repeat over and over again like this

c(0,30,130,200,230,300,330,400,430,500,0,30,130,200,230,300,330,400,430,500...

so for every 1 value in the first column I have a corresponding time in the second column then when I get to the 2's the times start over and each 2 has a corresponding time,

occasionally I will come across

c(3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4...

c(0,30,130,200,230,330,400,430,500,0,30,130,200,230,300,330,400,430,500...

Here one of the 3's is missing and the corresponding time of 300 is missing with it.

How can I go through my entire data frame and add these missing values? I need a way for R to go through and identify any missing values then insert a row and put the appropriate value, 1 to 365, in column one and the appropriate time with it. So for the given example R would add a row in between 230 and 330 and then place a 3 in the first column and 300 in the second. There are parts of the column that are missing several consecutive values. It is not just one here and there

Answer 1

EDIT: Solution with all 10 times clearly specified in advance and code tidy up/commenting

You need to create another data.frame containing every possible row and then merge it with your data.frame . The key aspect is the all.x = TRUE in the final merge which forces the gaps in your data to be highlighted. I simulated the gaps by sampling only 15 of the first 20 possible day/time combinations in your.dat

# create vectors for the days and times
the.days    = 1:365
the.times   = c(0,30,100,130,200,230,330,400,430,500)   # the 10 times to repeat

# create a master data.frame with all the times repeated for each day, taking only the first 20 observations
dat.all = data.frame(x1=rep(the.days, each=10), x2 = rep(the.times,times = 365))[1:20,]

# mimic your data.frame with some gaps in it (only 15 of 20 observations are present)
your.sample = sample(1:20, 15)
your.dat = data.frame(x1=rep(the.days, each=10), x2 = rep(the.times,times = 365), x3 = rnorm(365*10))[your.sample,]

# left outer join merge to include ALL of the master set and all of your matching subset, filling blanks with NA
merge(dat.all, your.dat, all.x = TRUE)

Here is the output from the merge, showing all 20 possible records with the gaps clearly visible as NA :

   x1  x2          x3
1   1   0          NA
2   1  30  1.23128294
3   1 100  0.95806838
4   1 130  2.27075361
5   1 200  0.45347199
6   1 230 -1.61945983
7   1 330          NA
8   1 400 -0.98702883
9   1 430          NA
10  1 500  0.09342522
11  2   0  0.44340164
12  2  30  0.61114408
13  2 100  0.94592127
14  2 130  0.48916825
15  2 200  0.48850478
16  2 230          NA
17  2 330  0.52789171
18  2 400 -0.16939587
19  2 430  0.20961745
20  2 500          NA

Answer 2

Here are a few NA handling functions that could help you getting started. For the inserting task, you should provide your own data using dput or a reproducible example.

df <- data.frame(x = sample(c(1, 2, 3, 4), 100, replace = T), 
                 y = sample(c(0,30,130,200,230,300,330,400,430,500), 100, replace = T))

nas <- sample(NA, 20, replace = T)
df[1:20, 1] <- nas
df$y <- ifelse(df$y == 0, NA, df$y)

# Columns x and y have NA's in diferent places.

# Logical test for NA
is.na(df)

# Keep not NA cases of one colum
df[!is.na(df$x),]
df[!is.na(df$y),]

# Returns complete cases on both rows
df[complete.cases(df),]

# Gives the cases that are incomplete.
df[!complete.cases(df),]

# Returns the cases without NAs
na.omit(df)