read.table，read.csv或扫描以读取R中的文本文件？

Question

我很困惑应该使用以下哪个？ （实际上到目前为止，所有这些都给我错误）：

> beef = read.csv("beef.txt", header = TRUE)
Error in read.table(file = file, header = header, sep = sep, quote = quote,  : 
  more columns than column names
> beef = scan("beef.txt")
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,  : 
  scan() expected 'a real', got '%'
> beef=read.table("beef.txt", header = FALSE, sep = " ")
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,  : 
  line 1 did not have 8 elements
> beef=read.table("beef.txt", header = TRUE, sep = " ")
Error in read.table("beef.txt", header = TRUE, sep = " ") : 
  more columns than column names

这是beef.txt文件的顶部，其余部分非常相似。

% http://lib.stat.cmu.edu/DASL/Datafiles/agecondat.html
% 
% Datafile Name: Agricultural Economics Studies
% Datafile Subjects: Agriculture , Economics , Consumer
% Story Names: Agricultural Economics Studies
% Reference: F.B. Waugh, Graphic Analysis in Agricultural Economics,
%   Agricultural Handbook No. 128, U.S. Department of Agriculture, 1957.
% Authorization: free use
% Description: Price and consumption per capita of beef and pork
%   annually from 1925 to 1941 together with other variables relevant to
%   an economic analysis of price and/or consumption of beef and pork
%   over the period.
% Number of cases: 17
% Variable Names:
% 
%   PBE = Price of beef (cents/lb)
%   CBE = Consumption of beef per capita (lbs)
%   PPO = Price of pork (cents/lb)
%   CPO = Consumption of pork per capita (lbs)
%   PFO = Retail food price index (1947-1949 = 100)
%   DINC = Disposable income per capita index (1947-1949 = 100)
%   CFO = Food consumption per capita index (1947-1949 = 100)
%   RDINC = Index of real disposable income per capita (1947-1949 = 100)
%   RFP = Retail food price index adjusted by the CPI (1947-1949 = 100)
% 
% The Data:
YEAR    PBE CBE PPO CPO PFO DINC    CFO RDINC   RFP
1925    59.7    58.6    60.5    65.8    65.8    51.4    90.9    68.5    877
1926    59.7    59.4    63.3    63.3    68  52.6    92.1    69.6    899
1927    63  53.7    59.9    66.8    65.5    52.1    90.9    70.2    883
1928    71  48.1    56.3    69.9    64.8    52.7    90.9    71.9    884
1929    71  49  55  68.7    65.6    55.1    91.1    75.2    895

当我使用fread时，数据存储非常奇怪，如下所示，您知道如何将其格式化为预期的格式吗？

> library(data.table)
> beef=fread("beef.txt", header = T, sep = " ")
> beef
    YEAR V2 V3 V4
 1: 1925 NA NA NA
 2: 1926 NA NA NA
 3: 1927 NA NA NA
 4: 1928 NA NA NA
 5: 1929 NA NA NA
 6: 1930 NA NA NA
 7: 1931 NA NA NA
 8: 1932 NA NA NA
 9: 1933 NA NA NA
10: 1934 NA NA NA
11: 1935 NA NA NA
12: 1936 NA NA NA
13: 1937 NA NA NA
14: 1938 NA NA NA
15: 1939 NA NA NA
16: 1940 NA NA NA
17: 1941 NA NA NA
         PBE\tCBE\tPPO\tCPO\tPFO\tDINC\tCFO\tRDINC\tRFP
 1: 59.7\t58.6\t60.5\t65.8\t65.8\t51.4\t90.9\t68.5\t877
 2:   59.7\t59.4\t63.3\t63.3\t68\t52.6\t92.1\t69.6\t899
 3:   63\t53.7\t59.9\t66.8\t65.5\t52.1\t90.9\t70.2\t883
 4:   71\t48.1\t56.3\t69.9\t64.8\t52.7\t90.9\t71.9\t884
 5:       71\t49\t55\t68.7\t65.6\t55.1\t91.1\t75.2\t895
 6: 74.2\t48.2\t59.6\t66.1\t62.4\t48.8\t90.7\t68.3\t874
 7:       72.1\t47.9\t57\t67.4\t51.4\t41.5\t90\t64\t791
 8:     79\t46\t49.5\t69.7\t42.8\t31.4\t87.8\t53.9\t733
 9:   73.1\t50.8\t47.3\t68.7\t41.6\t29.4\t88\t53.2\t752
10:   70.2\t55.2\t56.6\t62.2\t46.4\t33.2\t89.1\t58\t811
11:   82.2\t52.2\t73.9\t47.7\t49.7\t37\t87.3\t63.2\t847
12: 68.4\t57.3\t64.4\t54.4\t50.1\t41.8\t90.5\t70.5\t845
13:     73\t54.4\t62.2\t55\t52.1\t44.5\t90.4\t72.5\t849
14: 70.2\t53.6\t59.9\t57.4\t48.4\t40.8\t90.6\t67.8\t803
15:   67.8\t53.9\t51\t63.9\t47.1\t43.5\t93.8\t73.2\t793
16: 63.4\t54.2\t41.5\t72.4\t47.8\t46.5\t95.5\t77.6\t798
17:     56\t60\t43.9\t67.4\t52.2\t56.3\t97.5\t89.5\t830

当我按照评论中的说明阅读read.table时，我收到奇怪的输出（我没有像预期的那样阅读整齐）：

> beef=read.table("beef.txt", header = TRUE, sep = " ", comment.char="%")
> beef
   YEAR  X X.1 X.2
1  1925 NA  NA  NA
2  1926 NA  NA  NA
3  1927 NA  NA  NA
4  1928 NA  NA  NA
5  1929 NA  NA  NA
6  1930 NA  NA  NA
7  1931 NA  NA  NA
8  1932 NA  NA  NA
9  1933 NA  NA  NA
10 1934 NA  NA  NA
11 1935 NA  NA  NA
12 1936 NA  NA  NA
13 1937 NA  NA  NA
14 1938 NA  NA  NA
15 1939 NA  NA  NA
16 1940 NA  NA  NA
17 1941 NA  NA  NA
                PBE.CBE.PPO.CPO.PFO.DINC.CFO.RDINC.RFP
1  59.7\t58.6\t60.5\t65.8\t65.8\t51.4\t90.9\t68.5\t877
2    59.7\t59.4\t63.3\t63.3\t68\t52.6\t92.1\t69.6\t899
3    63\t53.7\t59.9\t66.8\t65.5\t52.1\t90.9\t70.2\t883
4    71\t48.1\t56.3\t69.9\t64.8\t52.7\t90.9\t71.9\t884
5        71\t49\t55\t68.7\t65.6\t55.1\t91.1\t75.2\t895
6  74.2\t48.2\t59.6\t66.1\t62.4\t48.8\t90.7\t68.3\t874
7        72.1\t47.9\t57\t67.4\t51.4\t41.5\t90\t64\t791
8      79\t46\t49.5\t69.7\t42.8\t31.4\t87.8\t53.9\t733
9    73.1\t50.8\t47.3\t68.7\t41.6\t29.4\t88\t53.2\t752
10   70.2\t55.2\t56.6\t62.2\t46.4\t33.2\t89.1\t58\t811
11   82.2\t52.2\t73.9\t47.7\t49.7\t37\t87.3\t63.2\t847
12 68.4\t57.3\t64.4\t54.4\t50.1\t41.8\t90.5\t70.5\t845
13     73\t54.4\t62.2\t55\t52.1\t44.5\t90.4\t72.5\t849
14 70.2\t53.6\t59.9\t57.4\t48.4\t40.8\t90.6\t67.8\t803
15   67.8\t53.9\t51\t63.9\t47.1\t43.5\t93.8\t73.2\t793
16 63.4\t54.2\t41.5\t72.4\t47.8\t46.5\t95.5\t77.6\t798
17     56\t60\t43.9\t67.4\t52.2\t56.3\t97.5\t89.5\t830

因此，感谢评论，分隔的不是空格而是制表符。 这是正确的答案：

> beef=read.table("beef.txt", header = TRUE, sep = "\t", comment.char="%")
> beef
    YEAR....PBE  CBE  PPO  CPO  PFO DINC  CFO RDINC RFP
1  1925    59.7 58.6 60.5 65.8 65.8 51.4 90.9  68.5 877
2  1926    59.7 59.4 63.3 63.3 68.0 52.6 92.1  69.6 899
3    1927    63 53.7 59.9 66.8 65.5 52.1 90.9  70.2 883
4    1928    71 48.1 56.3 69.9 64.8 52.7 90.9  71.9 884
5    1929    71 49.0 55.0 68.7 65.6 55.1 91.1  75.2 895
6  1930    74.2 48.2 59.6 66.1 62.4 48.8 90.7  68.3 874
7  1931    72.1 47.9 57.0 67.4 51.4 41.5 90.0  64.0 791
8    1932    79 46.0 49.5 69.7 42.8 31.4 87.8  53.9 733
9  1933    73.1 50.8 47.3 68.7 41.6 29.4 88.0  53.2 752
10 1934    70.2 55.2 56.6 62.2 46.4 33.2 89.1  58.0 811
11 1935    82.2 52.2 73.9 47.7 49.7 37.0 87.3  63.2 847
12 1936    68.4 57.3 64.4 54.4 50.1 41.8 90.5  70.5 845
13   1937    73 54.4 62.2 55.0 52.1 44.5 90.4  72.5 849
14 1938    70.2 53.6 59.9 57.4 48.4 40.8 90.6  67.8 803
15 1939    67.8 53.9 51.0 63.9 47.1 43.5 93.8  73.2 793
16 1940    63.4 54.2 41.5 72.4 47.8 46.5 95.5  77.6 798
17   1941    56 60.0 43.9 67.4 52.2 56.3 97.5  89.5 830

Answer 1

beef=read.table("beef.txt", header = TRUE, sep = " ", comment.char="%")

beef=read.table("beef.txt", header = TRUE, sep = "\t", comment.char="%") #after update

Answer 2

最近有一篇关于它的博客文章显示， fread是最快的，其余的都是一样的。 链接： http ： //statcompute.wordpress.com/2014/02/11/efficiency-of-importing-large-csv-files-in-r/

对于您而言，这并不重要，请使用您觉得最舒服的一种。

以下是使用fread的示例（假设使用TAB分隔符）：

library(data.table)
a = fread("data.csv", skip=26)
a
   YEAR  PBE  CBE  PPO  CPO  PFO DINC  CFO RDINC RFP
1: 1925 59.7 58.6 60.5 65.8 65.8 51.4 90.9  68.5 877
2: 1926 59.7 59.4 63.3 63.3 68.0 52.6 92.1  69.6 899
3: 1927 63.0 53.7 59.9 66.8 65.5 52.1 90.9  70.2 883
4: 1928 71.0 48.1 56.3 69.9 64.8 52.7 90.9  71.9 884
5: 1929 71.0 49.0 55.0 68.7 65.6 55.1 91.1  75.2 895

Answer 3

这是使用readLines基础替代方案。 这种方法要复杂得多，但是会返回准备好进行分析的数字数据。 但是，您必须手动计算原始数据文件中的列，然后再重新分配列名。

编辑

在底部，我添加了一个通用版本，该版本不需要手动计算列或手动添加列名。

请注意，无论数据是用空格还是制表符定界，这两个版本均适用。

这是原始版本的代码：

my.data <- readLines('c:/users/mmiller21/simple R programs/beef.txt')

ncols <- 10

header.info <- ifelse(substr(my.data, 1, 1) == '%', 1, 0)

my.data2 <- my.data[header.info==0]

my.data3 <- data.frame(matrix(unlist(strsplit(my.data2[-1], "[^0-9,.]+")), ncol=ncols, byrow=TRUE), stringsAsFactors = FALSE)

my.data4 <- as.data.frame(apply(my.data3, 2, function(x) as.numeric(x)))

colnames(my.data4) <- c('YEAR', 'PBE', 'CBE', 'PPO', 'CPO', 'PFO', 'DINC', 'CFO', 'RDINC', 'RFP')

> my.data4
     YEAR  PBE  CBE  PPO  CPO  PFO DINC  CFO RDINC RFP
[1,] 1925 59.7 58.6 60.5 65.8 65.8 51.4 90.9  68.5 877
[2,] 1926 59.7 59.4 63.3 63.3 68.0 52.6 92.1  69.6 899
[3,] 1927 63.0 53.7 59.9 66.8 65.5 52.1 90.9  70.2 883
[4,] 1928 71.0 48.1 56.3 69.9 64.8 52.7 90.9  71.9 884
[5,] 1929 71.0 49.0 55.0 68.7 65.6 55.1 91.1  75.2 895

以下是原始数据文件的内容：

% http://lib.stat.cmu.edu/DASL/Datafiles/agecondat.html
% 
% Datafile Name: Agricultural Economics Studies
% Datafile Subjects: Agriculture , Economics , Consumer
% Story Names: Agricultural Economics Studies
% Reference: F.B. Waugh, Graphic Analysis in Agricultural Economics,
%   Agricultural Handbook No. 128, U.S. Department of Agriculture, 1957.
% Authorization: free use
% Description: Price and consumption per capita of beef and pork
%   annually from 1925 to 1941 together with other variables relevant to
%   an economic analysis of price and/or consumption of beef and pork
%   over the period.
% Number of cases: 17
% Variable Names:
% 
%   PBE = Price of beef (cents/lb)
%   CBE = Consumption of beef per capita (lbs)
%   PPO = Price of pork (cents/lb)
%   CPO = Consumption of pork per capita (lbs)
%   PFO = Retail food price index (1947-1949 = 100)
%   DINC = Disposable income per capita index (1947-1949 = 100)
%   CFO = Food consumption per capita index (1947-1949 = 100)
%   RDINC = Index of real disposable income per capita (1947-1949 = 100)
%   RFP = Retail food price index adjusted by the CPI (1947-1949 = 100)
% 
% The Data:
YEAR    PBE CBE PPO CPO PFO DINC    CFO RDINC   RFP
1925    59.7    58.6    60.5    65.8    65.8    51.4    90.9    68.5    877
1926    59.7    59.4    63.3    63.3    68  52.6    92.1    69.6    899
1927    63  53.7    59.9    66.8    65.5    52.1    90.9    70.2    883
1928    71  48.1    56.3    69.9    64.8    52.7    90.9    71.9    884
1929    71  49  55  68.7    65.6    55.1    91.1    75.2    895

这是通用版本的代码：

my.data <- readLines('c:/users/mmiller21/simple R programs/beef.txt')

header.info <- ifelse(substr(my.data, 1, 1) == '%', 1, 0)

my.data2 <- my.data[header.info==0]

ncols <- length(read.table(textConnection(my.data2[1])))

my.data3 <- data.frame(matrix(unlist(strsplit(my.data2[-1], "[^0-9,.]+")), ncol=ncols, byrow=TRUE), stringsAsFactors = FALSE)

my.data4 <- as.data.frame(apply(my.data3, 2, function(x) as.numeric(x)))

#colnames(my.data4) <- c('YEAR', 'PBE', 'CBE', 'PPO', 'CPO', 'PFO', 'DINC', 'CFO', 'RDINC', 'RFP')
#my.data4

colnames(my.data4) <- read.table(textConnection(my.data2[1]), colClasses = c('character'))
my.data4

colSums(my.data4)

sum(my.data4$PPO)

Answer 4

正确答案如下：

beef=read.table("beef.txt", header = TRUE, sep = "", comment.char="%")