我就快到了。我想取一个字符串向量,并应用以下条件再次返回一个向量。在第三步中,我在遍历列表时遇到问题。
数据样本:
> assignment
[1] "female" "female"
[3] "female" "female"
[5] "male" "female, female"
[7] "female" "female, female, female"
[9] "female" "female"
[11] "female" "female"
[13] "female" "female"
[15] "female" "female"
[17] "female, female, female, female, female, male" "male"
[19] "female" NA
我所做的:
#Split each row into multiple strings
genders <- str_split(assignment, ", ")
#Assign 1 for female and 0 for male
females_count <- lapply(genders, str_count, pattern="female")
#Count how many women are in each list element
females_sum <- lapply(females_count, sum)
这里是我被卡住的地方,我想创建另一个向量,它的长度与
assignment
根据以下规则:
compare_genders <- function(females_count,females_sum){
ifelse (females_sum==1,
"single_female",
ifelse(females_sum>1&&females_count!=0,
"mult_males",
ifelse(females_sum==0&&length(females_count)==1,
"single_male",
ifelse(females_sum==0&&length(females_count)>1,
"mult_males",
ifelse(females_sum>1&&length(females_count)>1,
"mixed_genders","strange"))))
}
compare_genders(females_count = females_count,females_sum = females_sum)
但随后我得到一条错误消息:
Error in ifelse(females_sum[[i]] == 1, "single_female", ifelse(females_sum[[i]] > :
unused arguments ("mult_males", ifelse(females_sum[[i]] == 0 && length(females_count[[i]]) == 1, "single_male", ifelse(females_sum[[i]] == 0 && length(females_count[[i]]) > 1, "mult_males", ifelse(females_sum[[i]] > 1 && length(females_count[[i]]) > 1, "mixed_genders", "strange"))))
所需向量的示例:
> new_vector
[1] "single_female" "single_female"
[3] "single_female" "single_female"
[5] "single_male" "mult_females"
[7] "females"
"mult_females"
[9] "single_female" "single_female"
[11] "single_female" "single_female"
[13] "single_female" "single_female"
[15] "single_female" "single_female"
[17] "mixed_genders"
"single_male"
[19] "single_female" NA
