我想创建一个数据框，其中包含类＆＃34;字符＆＃34;的对象内的元素。
9 15 22 23 0 1,052 393
10 16 23 0 1 1,652 291
11 17 0 1 2 1,593 228
12 18 1 2 3 1,097 170
我一直试图将所有这些信息分成7列但没有成功。
以下是我使用的代码：
download.file("https://www.asx.com.au/documents/products/asx-24-market-dynamics-mar-18.pdf", "asx-24-market-dynamics-mar-18.pdf", mode = "wb")
#install.packages("pdftools")
library(pdftools)
txt <- pdf_text("asx-24-market-dynamics-mar-18.pdf")
# get second page text
page_2 <- txt[2]
# separate lines
library(dplyr)
library(stringr)
page_2a <- page_2 %>%
str_split(pattern = "\n") %>%
unlist()
# create "two tables"
tbl1 <- page_2a[6:29]
tbl2 <- page_2a[33:56]
# transform into a data frame
tbl1 <- ?
答案 0 :(得分：0)
我们可以先创建一个通用模式，然后使用separate生成所需的输出
library(stringr)
library(dplyr)
data$A <- str_replace(data$A, ",","")
data %>%
separate(A, c("a1", "a2", "a3", "a4","a5", "a6","a7", "a8"), " ")
数据
data <- read.table(text = "A
'9 15 22 23 0 1,052 393'
'10 16 23 0 1 1,652 291'
",header=T)
<强>更新
tbl1 <- str_replace(tbl1, "\r","") #replace \r at the end of line with ""
tbl1 <- trimws(tbl1,which = c("left")) #trim the white space where line start
tbl1 <- gsub("[[:blank:]]+", ":", tbl1) #replace any number of space with :
test_tbl1 <- data.frame(tbl1)
tbl1_final <- test_tbl1 %>%
separate(tbl1, c("Chicago*","London*","HK/Sing","Tokyo","Aust\n(AEST)","ASX\nSPI 20\U2122","90 Day","3 Year","10 Year","ASX\nSPI 200\U2122"
,"90 Day","3 Year","10 Year"), ":")
答案 1 :(得分：0)
library(pdftools); library(tidyverse)
现在您有两个以空格作为分隔符的字符向量，因此请使用read.table并传递给text参数：
head( read.table( text=tbl1 ) )
#---------
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
1 9 15 22 23 0 1,052 393 6,119 8,176 14 10,196 1,279 646
2 10 16 23 0 1 1,652 291 5,222 7,325 11 10,126 1,053 600
3 11 17 0 1 2 1,593 228 4,750 6,917 13 9,754 923 617
4 12 18 1 2 3 1,097 170 4,193 4,801 12 9,937 910 549
5 13 19 2 3 4 644 624 1,733 2,605 12 10,259 851 512
6 14 20 3 4 5 658 103 1,910 2,081 11 10,070 808 421
由于您似乎对此很陌生，因此当您在看似数字值的逗号中看到逗号时，我建议您非常怀疑。 R不使用逗号作为千位分隔符，因此列：V6，V8，V9，V11和V12至少是因子列。 （对于V13也适用，如str所示）
str( read.table( text=tbl1 ) )
'data.frame': 24 obs. of 13 variables:
$ V1 : int 9 10 11 12 13 14 15 16 17 18 ...
$ V2 : int 15 16 17 18 19 20 21 22 23 0 ...
$ V3 : int 22 23 0 1 2 3 4 5 6 7 ...
$ V4 : int 23 0 1 2 3 4 5 6 7 8 ...
$ V5 : int 0 1 2 3 4 5 6 7 8 9 ...
$ V6 : Factor w/ 24 levels "1,052","1,097",..: 1 5 4 2 16 17 21 20 24 3 ...
$ V7 : Factor w/ 24 levels "1,324","1,593",..: 19 15 13 7 22 4 21 20 3 9 ...
$ V8 : Factor w/ 24 levels "1,323","1,733",..: 24 22 21 18 2 3 11 1 4 6 ...
$ V9 : Factor w/ 24 levels "1,222","10,959",..: 22 21 19 12 9 7 8 1 16 18 ...
$ V10: int 14 11 13 12 12 11 9 10 0 13 ...
$ V11: Factor w/ 24 levels "10,070","10,119",..: 5 3 21 22 7 1 19 20 23 6 ...
$ V12: Factor w/ 24 levels "1,053","1,212",..: 4 1 24 23 22 19 20 21 10 11 ...
$ V13: Factor w/ 24 levels "1,009","1,019",..: 16 14 15 12 11 10 9 8 13 17 ...
您可以使用gsub修复大多数但不是所有列：
str( read.table( text=gsub(",","",tbl1) ) )
'data.frame': 24 obs. of 13 variables:
$ V1 : int 9 10 11 12 13 14 15 16 17 18 ...
$ V2 : int 15 16 17 18 19 20 21 22 23 0 ...
$ V3 : int 22 23 0 1 2 3 4 5 6 7 ...
$ V4 : int 23 0 1 2 3 4 5 6 7 8 ...
$ V5 : int 0 1 2 3 4 5 6 7 8 9 ...
$ V6 : Factor w/ 24 levels "1052","1097",..: 1 5 4 2 14 16 21 20 24 3 ...
$ V7 : int 393 291 228 170 624 103 525 44 1690 2308 ...
$ V8 : int 6119 5222 4750 4193 1733 1910 2116 1323 10606 12539 ...
$ V9 : int 8176 7325 6917 4801 2605 2081 2486 1222 5319 6119 ...
$ V10: int 14 11 13 12 12 11 9 10 0 13 ...
$ V11: int 10196 10126 9754 9937 10259 10070 9465 9606 9960 10239 ...
$ V12: int 1279 1053 923 910 851 808 839 843 1714 1874 ...
$ V13: int 646 600 617 549 512 421 316 297 585 688 ...
table(read.table( text=gsub(",","",tbl1) )[6])
1052 1097 1368 1593 1652 3323 3773 4054 4243 4648 565 599 608 644 6573 658 684 6966
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
782 814 823 863 875 Closed
1 1 1 1 1 1