Question

我试图将海量数据转换为数据库，但要使其有序化似乎太麻烦了。

是示例数据。

structure(list(V1 = structure(c(2L, 1L, 1L, 3L, 1L, 4L, 1L, 1L, 
1L, 5L), .Label = c("", "01.01 ? 01.06                          ", 
"02.01 ? 02.10                          ", "03.01 ? 03.07                          ", 
"Chapter 4 (04.01~04.10)"), class = "factor"), V2 = structure(c(2L, 
1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 3L), .Label = c("", "CTH", "WO"
), class = "factor")), .Names = c("V1", "V2"), class = "data.frame", row.names = c(NA, 
-10L))

我想要显示的结果就像这样

structure(list(V1 = c(1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 2.01, 
2.02, 2.03, 2.04, 2.05, 2.06, 2.07, 2.08, 2.09, 2.1, 3.01, 3.02, 
3.03, 3.04, 3.05, 3.06, 3.07, 4.01, 4.02, 4.03, 4.04, 4.05, 4.06, 
4.07, 4.08, 4.09, 4.1), V2 = structure(c(1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("CTH", 
"WO"), class = "factor")), .Names = c("V1", "V2"), class = "data.frame", row.names = c(NA, 
-33L))

你可以发现A行的每个单元格都包含一个间隔值。它将是我想要制作的数据库中几个单元格的缩写。

我应该将这些单元格标记为C行的相同代码 - 例如，01.01 - CTH的单元格，01.02 - CTH的单元格，01.03 - CTH的单元格，依此类推到0.06。

但是有几个问题。

如何将区间值的单元格划分为2个变量？
并根据间隔的第一个数据的值标记相同的代码。
数据的分类指标是章节，标题和副标题。
我想如果区间值的单元格只是说第一章，我想将所有子类别数据标记为与区间值对应的代码。并在新的csv上组织数据框。

让我知道如何用R来解决这些问题。

谢谢大家

Answer 1

它并不是特别优雅，但它可以完成工作。

library(dplyr)
library(stringr)
library(purrr)
library(tidyr)
input %>% 
  filter(V1 != "") %>% 
  mutate(lim = str_extract_all(V1, "\\d{2}\\.\\d{2}"),
         lim = map_chr(lim, paste0, collapse = ",")) %>% 
  separate(lim, 
           into = c("low", "high"),
           sep = ",") %>% 
  separate(low,
           into = c("prefix", "low"),
           sep = "\\.") %>% 
  mutate(high = str_replace(hight, "^.+\\.", ""),
         id = row_number()) %>% 
  select(id, V2, prefix, low, high) %>% 
  split(f = list(.$id)) %>% 
  map(.f = function(rdf){
    suffix <- (as.numeric(rdf$low):as.numeric(rdf$high)) / 100
    data.frame(V1 = as.numeric(rdf$prefix) + suffix,
               V2 = rep(rdf$V2, length(suffix)),
               stringsAsFactors = FALSE)
  }) %>% 
  bind_rows()

Answer 2

这是一个混合基础R-stringr - data.table方法。

library(stringr)
library(data.table)
aa = aa = structure(list(V1 = structure(c(2L, 1L, 1L, 3L, 1L, 4L, 1L, 1L, 1L, 5L), 
                          .Label = c("", "01.01 ? 01.06                          ", "02.01 ? 02.10                          ", "03.01 ? 03.07                          ", 
                                                "Chapter 4 (04.01~04.10)"), class = "factor"), 
           V2 = structure(c(2L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 3L),
                          .Label = c("", "CTH", "WO"), class = "factor")), .Names = c("V1", "V2"),
      class = "data.frame", row.names = c(NA, -10L))
aa = aa[aa$V1 !="",]
a1 = str_trim(aa$V1)

cc  = str_split_fixed(aa$V1, "\\?|~", 2)
cc2 = apply(cc, 2, str_extract_all, "\\d+\\.\\d+", simplify = T)
cc2 = apply(cc2, 2, as.numeric)
cc2 = cbind(cc2, as.character(aa$V2))


range_col = list()
for(i in 1:nrow(cc2)){
  range_col[[i]] = data.table(V1 = seq(cc2[i,1], cc2[i,2], by = 0.01),
                              V2 = cc2[i, 3])
}
rr = rbindlist(range_col)

导致

      V1  V2
 1: 1.01 CTH
 2: 1.02 CTH
 3: 1.03 CTH
 4: 1.04 CTH
 5: 1.05 CTH
 6: 1.06 CTH
 7: 2.01 CTH
 8: 2.02 CTH
 9: 2.03 CTH
10: 2.04 CTH
11: 2.05 CTH
12: 2.06 CTH
13: 2.07 CTH
14: 2.08 CTH
15: 2.09 CTH
16: 2.10 CTH
17: 3.01 CTH
18: 3.02 CTH
19: 3.03 CTH
20: 3.04 CTH
21: 3.05 CTH
22: 3.06 CTH
23: 3.07 CTH
24: 4.01  WO
25: 4.02  WO
26: 4.03  WO
27: 4.04  WO
28: 4.05  WO
29: 4.06  WO
30: 4.07  WO
31: 4.08  WO
32: 4.09  WO
33: 4.10  WO

Answer 3

structure(list(V1 = structure(c(2L, 1L, 1L, 3L, 1L, 4L, 1L, 1L, 
1L, 5L), .Label = c("", "01.01 ? 01.06                          ", 
"02.01 ? 02.10                          ", "03.01 ? 03.07                          ", 
"Chapter 4 (04.01~04.10)"), class = "factor"), V2 = structure(c(2L, 
1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 3L), .Label = c("", "CTH", "WO"
), class = "factor")), .Names = c("V1", "V2"), class = "data.frame", row.names = c(NA, -10L)) -> df1

structure(list(V1 = c(1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 2.01, 
2.02, 2.03, 2.04, 2.05, 2.06, 2.07, 2.08, 2.09, 2.1, 3.01, 3.02, 
3.03, 3.04, 3.05, 3.06, 3.07, 4.01, 4.02, 4.03, 4.04, 4.05, 4.06, 
4.07, 4.08, 4.09, 4.1), V2 = structure(c(1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("CTH", 
"WO"), class = "factor")), .Names = c("V1", "V2"), class = "data.frame", row.names = c(NA, -33L)) -> df2


library(dplyr)
library(stringr)
library(purrr)
library(tidyr)


df1 %>% 
  filter(V1 != "") %>% 
  mutate(lim = str_extract_all(V1, "\\d{2}\\.\\d{2}"),
         lim = map_chr(lim, paste0, collapse = ",")) %>% 
  separate(lim, into = c("low", "high"), sep = ",", convert = T) %>%
  mutate(value = map2(low, high, ~paste0(seq(.x,.y, by=0.01), collapse = ","))) %>%
  select(V2, value) %>%
  separate_rows(value, convert = T)

Answer 4

这也说明了当您拥有一个数字时的情况......命名为您的数据df：

library(tidyverse)
as.character(df$V1) %>% 
  # extract all matching, ie 12.34
  str_extract_all(pattern = "\\d{2}\\.\\d{2}") %>% 
  # convert to matrix
  map(as.matrix) %>% 
  # transpose, such that it is only one column
  map(t) %>% 
  # convert to tibble (or data frame)
  map(as_tibble) %>% 
  # bind rows together, regardless of single or double rows
  reduce(bind_rows) %>% 
  # name columns
  set_names(nm = c("from", "to")) %>% 
  # if single values occur...
  mutate(to = ifelse(is.na(to), from, to),
         # add info of second column
         second = as.character(df$V2)) %>% 
  # drop missing rows
  drop_na() %>% 
  # convert to double for seq
  mutate_at(.vars = vars(from, to), as.double) %>% 
  group_by_all() %>% 
  # create list of individual length using seq
  mutate(first = list(seq(from = from, to = to, by = .01))) %>% 
  unnest() %>% ungroup() %>% 
  select(first, second)

如何根据R

4 个答案: