英文:
mutate() R replace -1 in column with the mean but all values are unconditionally replaced
问题
问题
尝试将列中的-1值用数据框列(HRS1)的mean()值进行填充或替换。我曾经使用了str_replace()，现在使用replace()来替代-1，该列是数值型的，但是当测试-1时，数据框列的所有值都会更改为mutate上的mean()。
这不是缺失值的填充/替代，而是需要替换 -1 值的列。
代码
mean_HRS1 = mean(df_survey$HRS1)
df_survey %>%
mutate(HRS1 = replace(HRS1, -1, mean_HRS1))
英文:
Problem
Just trying to impute or replace a -1 value in column with the mean() value of data.frame column (HRS1). I had str_replace() and now replace() for -1, the column is numeric, but when test for -1, then ALL values of the dataframe column are changed to mean() on mutate.
This is NOT a missing value imputation/replacement, but a column that has -1 values that need to be replaced.
Code
mean_HRS1 = mean(df_survey$HRS1)
df_survey %>%
mutate(HRS1 = replace(HRS1, -1, mean_HRS1))
答案1
得分: 1
你应该将逻辑向量 HRS1 == -1 或索引向量 which(HRS1 == -1) 传递给 replace 函数。
另一个问题是，在计算均值时应该排除 -1。给定 HRS1 为 c(6, 7, 8, -1)。mean(HRS1) 实际上是 (6+7+8-1) / 4，但你所需的应该是 (6+7+8) / 3。
df_survey %>%
mutate(HRS1 = replace(HRS1, HRS1 == -1, mean(HRS1[HRS1 != -1])))
英文:
You should pass an logical vector, HRS1 == -1, or an index vector, which(HRS1 == -1), into replace.
Another issue is that -1 should be excluded when computing the mean. Given HRS1 is c(6, 7, 8, -1). mean(HRS1) is actually (6+7+8-1) / 4, but what you need should be (6+7+8) / 3.
df_survey %>%
mutate(HRS1 = replace(HRS1, HRS1 == -1, mean(HRS1[HRS1 != -1])))
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