Interpolating missing values in timeframe ts()

The zoo package in R allows you to work with time series that have unevenly spaced time intervals, and variable frequencies (no need to interpolate data). The {mblm} package does a Thiel-Sen regression on time-series-like data. Looking at your dummy data I assume that you want to do the regression for each month. Try this:

a <- "
year	month	day	chickens
2000	1	2	1
2000	1	5	2
2000	1	8	3
2000	1	11	4
2000	1	14	5
2000	1	17	6
2000	1	20	7
2000	1	23	8
2000	1	26	9
2000	2	3	2
2000	2	6	4
2000	2	9	6
2000	2	12	8
2000	2	15	10
2000	2	18	12
2000	2	21	14
2000	2	24	16
2000	2	27	18
2000	3	1	3
2000	3	4	6
2000	3	7	9
2000	3	10	12
2000	3	13	15
2000	3	16	18
2000	3	19	21
2000	3	22	24
2000	3	25	27
2000	3	28	30
2000	4	3	4
2000	4	6	8
2000	4	9	12
2000	4	12	16
2000	4	15	20
2000	4	18	24
2000	4	21	28
2000	4	24	32
2000	4	27	36
2000	5	3	5
2000	5	6	10
2000	5	9	15
2000	5	12	20
2000	5	15	25
2000	5	18	30
2000	5	21	35
2000	5	24	40
2000	5	27	45
2000	6	3	6
2000	6	6	12
2000	6	9	18
2000	6	12	24
2000	6	15	30
2000	6	18	36
2000	6	21	42
2000	6	24	48
2000	6	27	54
2000	7	3	7
2000	7	6	14
2000	7	9	21
2000	7	12	28
2000	7	15	35
2000	7	18	42
2000	7	21	49
2000	7	24	56
2000	7	27	63
2000	8	3	8
2000	8	6	16
2000	8	9	24
2000	8	12	32
2000	8	15	40
2000	8	18	48
2000	8	21	56
2000	8	24	64
2000	8	27	72
2000	9	3	9
2000	9	6	18
2000	9	9	27
2000	9	12	36
2000	9	15	45
2000	9	18	54
2000	9	21	63
2000	9	24	72
2000	9	27	81
2000	10	3	10
2000	10	6	20
2000	10	9	30
2000	10	12	40
2000	10	15	50
2000	10	18	60
2000	10	21	70
2000	10	24	80
2000	10	27	90
2000	11	3	11
2000	11	6	22
2000	11	9	33
2000	11	12	44
2000	11	15	55
2000	11	18	66
2000	11	21	77
2000	11	24	88
2000	11	27	99
2000	12	6	12
2000	12	9	24
2000	12	12	36
2000	12	18	48
2000	12	21	60
2000	12	24	72
2000	12	27	84"

df <- read.delim(text=a, header=TRUE)

# making a date column
df$date <- as.Date(with(df, paste(year, month, day,sep="/")), "%Y/%m/%d")

library(zoo)
library(mblm)

for(ii in 1:12) {
  tmp <- df[df$month==ii,]
  print(tmp)
  zz <- zoo(tmp$chickens, tmp$date)
  plot(zz, type="b")
  fit <- mblm(chickens ~ day, dataframe=tmp)
  print(summary(fit))
}

It's a for() loop, I know, but I'll leave someone else to get a nice dplyr or lapply() version working.

HTH