计算字符串中第一个，第二个和第三个字的字符数

Question

我需要了解开发一个可以计算字符串中第二个和第三个字的字符数的代码。

我得到了这个代码，但它只适用于第一个单词的字符数。

现在我只允许使用Spark SQL或dplyr包。

这就是我在第一个单词中为count个字符所做的

INSTR(NAME_NORM_LONG,' ')-1)

预期的结果是计算字符并在新列中显示结果。

word="hey I am Scott"

characters_word1 | characters_word2 | characters_word3 

          3               1                   2

现在我运行此代码进行测试（尝试找到第二个单词）：

test_query<-test_query %>% 
mutate(Total_char=nchar(NAME_NORM_LONG))%>%
mutate(Name_has_numbers=str_detect(NAME_NORM_LONG,"[[:digit:]]"))%>%
mutate(number_words=LENGTH(NAME_NORM_LONG) - LENGTH(REPLACE(NAME_NORM_LONG, ' ', '')) + 1)%>%
mutate(number_chars_w1=INSTR(NAME_NORM_LONG,' ')-1)%>%
mutate(number_chars_w2=substr(NAME_NORM_LONG,number_chars_w1+1,LENGTH(NAME_NORM_LONG)))``` and the result I am having is this one ```test_query
# Source: spark<?> [?? x 7]
   PBIN0 NAME_NORM_LONG Total_char Name_has_numbers number_words number_chars_w1
   <int> <chr>               <int> <lgl>                   <dbl>           <dbl>
1 4.01e8 GM BUILDERS            11 FALSE                       2               2
# … with 1 more variable: number_chars_w2 <chr>
Warning messages:
1: In substr(NAME_NORM_LONG, number_chars_w1, 1) :
  NAs introduced by coercion
2: In substr(NAME_NORM_LONG, number_chars_w1, 1) :
  NAs introduced by coercion
3: In substr(NAME_NORM_LONG, number_chars_w1, 1) :
  NAs introduced by coercion
4: In substr(NAME_NORM_LONG, number_chars_w1, 1) :
  NAs introduced by coercion
5: In substr(NAME_NORM_LONG, number_chars_w1, 1) :
  NAs introduced by coercion```

Answer 1

使用str_split()怎么样？

word="hey I am Scott"

word_list = stringr::str_split(word, " ")

n = length(word_list[[1]])
for (i in 1:n){
  first_row = paste0("characters_word", 1:n)
  second_row = sapply(word_list[[1]], nchar)
}

df = data.frame(first_row, second_row)