I have an R dataframe where one of the columns is a factor whose levels have an implicit ordering. How can I convert the factor levels to specific integers in the following manner:
For example, here is my data frame:
agree <- c("Strongly agree", "Somewhat disagree", "Somewhat agree",
"Neutral", "Strongly agree", "Strongly disagree", "Neutral")
age <- c(41, 35, 29, 42, 31, 22, 58)
df <- data.frame(age, agree)
df
# age agree
# 1 41 Strongly agree
# 2 35 Somewhat disagree
# 3 29 Somewhat agree
# 4 42 Neutral
# 5 31 Strongly agree
# 6 22 Strongly disagree
# 7 58 Neutral
str(df)
# 'data.frame': 7 obs. of 2 variables:
# $ age : num 41 35 29 42 31 22 58
# $ agree: Factor w/ 5 levels "Neutral","Somewhat agree",..: 4 3 2 1 4 5 1
Now, I would like to convert the agree
column to be an integer column using the mapping that I showed above.
I already searched these other questions about converting factor to integer, but they do not related to maintaining the factor ordering.
"How to convert a factor to an integer\numeric without a loss of information?"
"Convert factor to integer"
"Convert factor to integer in a data frame"
You need to define the order of factors first:
ordering <- c("Strongly disagree", "Somewhat disagree", "Neutral", "Somewhat agree", "Strongly agree")
Then, when you first create your factor, you should use that definition:
agreeFactor <- factor(agree, levels = ordering)
Then, you should be able to get your ordered factor:
as.numeric(agreeFactor)
You can also just apply the order when using as.numeric(), but this can lead to inconsistencies if you decide to later retrieve your numeric vector and forget to apply the "levels = " argument.
e: If you want to directly import the numeric into your dataframe, simply use:
df$agree <- as.numeric(factor(df$agree, levels = ordering))
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