NaN 作为使用 tapply 的结果，在计算均值时

Question

查了很多，也有类似的问题，就是简单的一件事情都看不懂。我正在尝试计算不同棒球职位的平均工资。

library(Lahman)
library(tidyverse)

data("Fielding")
data(Salaries)

# First, I need to merge two datasets
merged.df <- merge(Fielding, Salaries, by = "playerID", na.rm = TRUE)
merged.df.2002 <- merged.df[merged.df$yearID.x == "2002",]

# Let's try tapply
mean.salary <- tapply(merged.df.2002$POS, merged.df.2002$salary, mean, na.rm = TRUE)
# So it gives me an error
# In mean.default(X[[i]], ...) :
#  argument is not numeric or logical: returning NA

class(merged.df.2002$POS)
class(merged.df.2002$salary)

# Very likely POS column is factor for some reason.
# Coerce them through 
merged.df.2002$POS <- as.numeric(as.character(merged.df.2002$POS))
# Warning message:
# NAs introduced by coercion 
merged.df.2002$salary <- as.numeric(as.character(merged.df.2002$salary))
#as.numeric(merged.df.2002$salary)
class(merged.df.2002$salary)

# Let's try tapply again
mean.salary <- tapply(merged.df.2002$POS, merged.df.2002$salary, mean, na.rm 
= TRUE)
mean.salary

60000   62500   63500   65000   67000   67500   68000   68750   70000   
71000   72500   77500   78000   80000   82000   82500 
NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN     
NaN     NaN     NaN     NaN     NaN     NaN

有什么想法吗？非常感谢！

Answer 1

好吧，这很容易，但我搞砸了

merged.df <- merge(Fielding, Salaries) 
# So, my mistake was that I merged only by playerid.

merged.df.2002 <- merged.df[merged.df$yearID == 2002, ] 
# we pick 2002 year from the merged dataset

# use tapply for mean
merged.df.mean <- tapply(merged.df.2002$salary, merged.df.2002$POS, mean, 
na.rm = TRUE)

#      1B      2B      3B       C      OF       P      SS 
# 2543845 1400543 1547836 1787933 2659230 2150887 1979732

NaN 作为使用 tapply 的结果，在计算均值时

NaN as a result of using tapply, when calculating means

r

tapply