R / tidyr::complete - 动态填充缺失值

Question

我正在使用 tidyr::complete() 在具有许多列的数据框中包含缺失的行，从而导致 NA 值。如果我没有明确的列名列表，如何指示 fill 选项将 NA 值替换为 0？

示例：

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)))

解决方法是使用 df.complete[is.na(df.complete)] <- 0，但这会带来替换太多值的危险。

Answer 1

这是一种首先重塑数据的方法：

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

R / tidyr::complete - 动态填充缺失值

R / tidyr::complete - filling missing values dynamically

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