1

我有一个这样的 .csv 文件：

+-------+---------+------+-------+
| CONN  |  TABLE  | COLS | OWNER |
+-------+---------+------+-------+
| ONE   | TABLE_A |   10 | MIKE  |
| ONE   | TABLE_B |    9 | MIKE  |
| ONE   | TAB_A   |   11 | KIM   |
| ONE   | TAB_B   |   14 | KIM   |
| TWO   | TABLE_A |    9 | MIKE  |
| TWO   | TABLE_B |    9 | MIKE  |
| TWO   | TAB_A   |   11 | KIM   |
| TWO   | TAB_D   |   56 | KIM   |
| THREE | TABLE_A |    9 | MIKE  |
| THREE | TABLE_C |    3 | MIKE  |
| THREE | TABLE_D |   11 | KIM   |
| THREE | TAB_A   |   11 | KIM   |
+-------+---------+------+-------+

我想按 conn 和 owner 比较表和 cols。如何重塑这些数据以进行比较？我的数据在这里：

dat <- structure(list(CONN = c("ONE", "ONE", "ONE", "ONE", "TWO", "TWO", 
"TWO", "TWO", "THREE", "THREE", "THREE", "THREE"), TABLE = c("TABLE_A", 
"TABLE_B", "TAB_A", "TAB_B", "TABLE_A", "TABLE_B", "TAB_A", "TAB_D", 
"TABLE_A", "TABLE_C", "TABLE_D", "TAB_A"), COLS = c(10L, 9L, 
11L, 14L, 9L, 9L, 11L, 56L, 9L, 3L, 11L, 11L), OWNER = c("MIKE", 
"MIKE", "KIM", "KIM", "MIKE", "MIKE", "KIM", "KIM", "MIKE", "MIKE", 
"KIM", "KIM")), .Names = c("CONN", "TABLE", "COLS", "OWNER"), class = "data.frame", row.names = c(NA, 
-12L))

我试过类似的东西：

reshape(dat, varying=c('TABLE', 'COLS'), v.names=C('CONN', 'OWNER'), direction='long')
Error in C("CONN", "OWNER") : object not interpretable as a factor
4

3 回答 3

4

所有这些 CAPITAL 列名都让您在 shift 键上有点粘 - 你做到了C('CONN','OWNER')。小写的c工作方式如下：

> reshape(dat, varying=c('TABLE', 'COLS'), v.names=c('CONN', 'OWNER'), direction='long')
      CONN OWNER time id
1  TABLE_A    10    1  1
2  TABLE_B     9    1  2
3    TAB_A    11    1  3
4    TAB_B    14    1  4
5  TABLE_A     9    1  5
6  TABLE_B     9    1  6
7    TAB_A    11    1  7
8    TAB_D    56    1  8
9  TABLE_A     9    1  9
10 TABLE_C     3    1 10
11 TABLE_D    11    1 11
12   TAB_A    11    1 12
于 2013-08-22T13:47:46.920 回答
1

我通常会发现该reshape2包更直观：只需将所需的行（相应的列）放在~.

dcast( CONN + OWNER ~ TABLE, data = dat, value.var="COLS" )
#    CONN OWNER TAB_A TAB_B TAB_D TABLE_A TABLE_B TABLE_C TABLE_D
# 1   ONE   KIM    11    14    NA      NA      NA      NA      NA
# 2   ONE  MIKE    NA    NA    NA      10       9      NA      NA
# 3 THREE   KIM    11    NA    NA      NA      NA      NA      11
# 4 THREE  MIKE    NA    NA    NA       9      NA       3      NA
# 5   TWO   KIM    11    NA    56      NA      NA      NA      NA
# 6   TWO  MIKE    NA    NA    NA       9       9      NA      NA
于 2013-08-22T14:09:53.393 回答
0

也可以使用包中的gather()功能：tidyr

Function:       gather(data, key, value, ..., na.rm = FALSE, convert = FALSE)
Same as:        data %>% gather(key, value, ..., na.rm = FALSE, convert = FALSE)

Arguments:
    data:           data frame
    key:            column name representing new variable
    value:          column name representing variable values
    ...:            names of columns to gather (or not gather)
    na.rm:          option to remove observations with missing values (represented by NAs)
    convert:        if TRUE will automatically convert values to logical, integer, numeric, complex or 
                    factor as appropriate
于 2016-03-14T13:09:22.373 回答