3

I'm using tidyr::complete() to include missing rows in a data frame with many columns, leading to NAs values. How can I instruct the fill option to replace the NA values with 0 if I don't have an explicit list of column names?

Example: 

df <- data.frame(year = c(2010, 2013:2015),
                 age.21 = runif(4, 0, 10),
                 age.22 = runif(4, 0, 10),
                 age.23 = runif(4, 0, 10),
                 age.24 = runif(4, 0, 10),
                 age.25 = runif(4, 0, 10))

# replaces missing values with NA - not what I want
df.complete <- complete(df, year = 2010:2015)

# replaces missing values with 0 - works, but needs explicit list
df.complete <- complete(df, year = 2010:2015, fill = list(age.21 = 0, age.22 = 0,
                                                          age.23 = 0, age.24 = 0,
                                                          age.25 = 0))


# throws error (is.list(replace) is not TRUE)
df.complete <- complete(df, year = 2010:2015, fill = 0)

# replaces missing values with NA - not what I want
df.complete <- complete(df, year = 2010:2015, fill = list(rep(0,6)))

A workaround could be to use df.complete[is.na(df.complete)] <- 0, but that bears the danger of replacing too many values.

Timm S.
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1 Answers1

3

Here's a way with reshaping the data first:

df %>% 
  gather("var", "val", -year) %>% 
  complete(year = 2010:2015, var, fill = list(val = 0)) %>%
  spread(var, val)

Source: local data frame [6 x 6]

   year   age.21   age.22    age.23   age.24    age.25
  (dbl)    (dbl)    (dbl)     (dbl)    (dbl)     (dbl)
1  2010 8.940997 7.787210 1.5747435 9.874449 5.2228670
2  2011 0.000000 0.000000 0.0000000 0.000000 0.0000000
3  2012 0.000000 0.000000 0.0000000 0.000000 0.0000000
4  2013 2.965928 6.495460 0.8966319 2.849262 0.2430174
5  2014 4.608676 1.946671 1.5765912 8.551907 0.3146824
6  2015 7.359407 4.414294 4.3419163 4.082509 1.5770299
erc
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