拆分字符串列以创建新的二进制列

Question

我的数据有一列,我正在尝试使用行中每个"/"之后的内容创建其他列.以下是数据的前几行:

> dput(mydata)
structure(list(ALL = structure(c(1L, 4L, 4L, 3L, 2L), .Label = c("/
ca/put/sent_1/fe.gr/eq2_on/eq2_off",
"/ca/put/sent_1/fe.gr/eq2_on/eq2_off/cbr_LBL", "/ca/put/sent_1/fe.g
r/eq2_on/eq2_off/cni_at.p3x.4",
"/ca/put/sent_1/fe.gr/eq2_on/eq2_off/hi.on/hi.ov"), class = "factor
")), .Names = "ALL", class = "data.frame", row.names = c(NA,
-5L))

如果变量出现在行中,结果应该如此(数据框)在新列中带有"1",否则为"0":

> dput(Result)
structure(list(ALL = structure(c(1L, 4L, 5L, 3L, 2L), .Label = c("/ca
/put/sent_1/fe.gr/eq2_on/eq2_off",
"/ca/put/sent_1/fe.gr/eq2_on/eq2_off/cbr_LBL", "/ca/put/sent_1/fe.gr/
eq2_on/eq2_off/cni_at.p3x.4",
"/ca/put/sent_1/fe.gr/eq2_on/eq2_off/hi.on/hi.ov", "/ca/put/sent_1fe.
gr/eq2_on/eq2_off/hi.on/hi.ov"
), class = "factor"), ca = c(1L, 1L, 1L, 1L, 1L), put = c(1L,
1L, 1L, 1L, 1L), sent_1 = c(1L, 1L, 1L, 1L, 1L), fe.gr = c(1L,
1L, 1L, 1L, 1L), eq2_on = c(1L, 1L, 1L, 1L, 1L), eq2_off = c(1L,
1L, 1L, 1L, 1L), hi.on = c(0L, 1L, 1L, 0L, 0L), hi.ov = c(0L,
1L, 1L, 0L, 0L), cni_at.p3x.4 = c(0L, 0L, 0L, 1L, 0L), cbr_LBL = c(0L
,
0L, 0L, 0L, 1L)), .Names = c("ALL", "ca", "put", "sent_1", "fe.gr",
"eq2_on", "eq2_off", "hi.on", "hi.ov", "cni_at.p3x.4", "cbr_LBL"
), class = "data.frame", row.names = c(NA, -5L))

我尝试了很多功能,包括strsplit和sapply:

sapply(strsplit(as.character(mydata$ALL), “\\/”), “[[“, 2) #returns "ca"s only

sapply(strsplit(as.character(mydata$ALL), "\\/"), "[[", 3) #returns "put"s only

有数百万行,我非常欣赏任何快速有效的内容.

Answer 1

mtabuate从我维护的qdapTools包中使用：

library(qdapTools)
mtabulate(strsplit(as.character(dat[[1]]), "/"))

##   V1 ca cbr_LBL cni_at.p3x.4 eq2_off eq2_on fe.gr hi.on hi.ov put sent_1 sent_1fe.gr
## 1  1  1       0            0       1      1     1     0     0   1      1           0
## 2  1  1       0            0       1      1     1     1     1   1      1           0
## 3  1  1       0            0       1      1     0     1     1   1      0           1
## 4  1  1       0            1       1      1     1     0     0   1      1           0
## 5  1  1       1            0       1      1     1     0     0   1      1           0

Answer 2

您可以cSplit_e从我的“splitstackshape”包中使用：

library(splitstackshape)
cSplit_e(mydata, "ALL", "/", type = "character", fill = 0)
#                                                ALL ALL_ca ALL_cbr_LBL
# 1              /ca/put/sent_1/fe.gr/eq2_on/eq2_off      1           0
# 2  /ca/put/sent_1/fe.gr/eq2_on/eq2_off/hi.on/hi.ov      1           0
# 3   /ca/put/sent_1fe.gr/eq2_on/eq2_off/hi.on/hi.ov      1           0
# 4 /ca/put/sent_1/fe.gr/eq2_on/eq2_off/cni_at.p3x.4      1           0
# 5      /ca/put/sent_1/fe.gr/eq2_on/eq2_off/cbr_LBL      1           1
#   ALL_cni_at.p3x.4 ALL_eq2_off ALL_eq2_on ALL_fe.gr ALL_hi.on ALL_hi.ov ALL_put
# 1                0           1          1         1         0         0       1
# 2                0           1          1         1         1         1       1
# 3                0           1          1         0         1         1       1
# 4                1           1          1         1         0         0       1
# 5                0           1          1         1         0         0       1
#   ALL_sent_1 ALL_sent_1fe.gr
# 1          1               0
# 2          1               0
# 3          0               1
# 4          1               0
# 5          1               0

（注意：我认为您的第 3 行存在问题，dput这就是为什么它与您想要的输出不匹配。请注意，第 3 行中的第三项是“sent_1fe.gr”，它们之间没有“/”。）