当一个正常的数据帧，如"test"下面,完全在十进制转换成"!",":
a <- c(1:34)
b <- rnorm(34, mean=33, sd=7)
test <- cbind.data.frame(a,b)
write.table(file="test.csv",test, row.names = F, dec=",", sep = ";")
我的其他数据帧没有出现"，"为小数。我猜"grep"的上游用法和";aggregate"某种程度上是转换的障碍。下面的Str输出，当我有五个变量时，说三个。我如何准备数据帧可访问的十进制转换?
Group.1 Group.2 x.mean x.sd x.cv
1 P1 Compound 1: IgG1 11.94520000 0.11435889 0.95736270
2 P2 Compound 1: IgG1 10.29220000 0.06536700 0.63511201
3 P1 Compound 2: IgG2 10.07450000 0.05682967 0.56409417
4 P2 Compound 2: IgG2 19.66320000 0.16354259 0.83171908
...
'data.frame': 12 obs. of 3 variables:
$ Group.1: Factor w/ 10 levels "","FBS","ID",..: 9 10 9 10 9 10 9 10 9 10 ...
$ Group.2: Factor w/ 11 levels "Compound 1: IgG1",..: 1 1 2 2 3 3 4 4 5 5 ...
$ x : num [1:12, 1:3] 11.95 10.29 10.07 19.66 4.21 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr "mean" "sd" "cv"
dput.
structure(list(Group.1 = structure(c(9L, 10L, 9L, 10L, 9L, 10L
), .Label = c("", "FBS", "ID", "K1", "K2", "K3", "K4", "K5",
"P1", "P2"), class = "factor"), Group.2 = structure(c(1L, 1L,
2L, 2L, 3L, 3L), .Label = c("Compound 1: IgG1", "Compound 2: IgG2",
"Compound 3: IgG3", "Compound 4: IgG3-723", "Compound 5: IgG4",
"Compound 6: Total-IgG", "Compound 7: IgG1_IS", "Compound 8: IgG2_IS",
"Compound 9: IgG3_IS", "Compound 10: IgG4_IS", "Compound 11: Total_IgG_IS"
), class = "factor"), x = structure(c(11.9452, 10.2922, 10.0745,
19.6632, 4.2135, 3.7465, 0.114358889272131, 0.0653669981293651,
0.0568296675259594, 0.163542587046242, 0.0569370997973496, 0.0253651116474753,
0.957362700265639, 0.63511200840797, 0.564094173665784, 0.831719084616146,
1.35130176331671, 0.677034876484061), .Dim = c(6L, 3L), .Dimnames = list(
NULL, c("mean", "sd", "cv")))), row.names = c(NA, 6L), class = "data.frame")
@r2evans的预感是对的。问题是test$x，它是一个矩阵，这是问题的根源。您得到的输出根本没有意义，也不能很好地表示数据。如果您将矩阵列直接包含到数据框架中，它将按预期工作。
test <- structure(
list(Group.1 = structure(
c(9L, 10L, 9L, 10L, 9L, 10L),
.Label = c("", "FBS", "ID", "K1", "K2", "K3", "K4", "K5", "P1", "P2"),
class = "factor"),
Group.2 = structure(
c(1L, 1L, 2L, 2L, 3L, 3L),
.Label = c("Compound 1: IgG1", "Compound 2: IgG2", "Compound 3: IgG3",
"Compound 4: IgG3-723", "Compound 5: IgG4", "Compound 6: Total-IgG",
"Compound 7: IgG1_IS", "Compound 8: IgG2_IS", "Compound 9: IgG3_IS",
"Compound 10: IgG4_IS", "Compound 11: Total_IgG_IS"),
class = "factor"),
x = structure(c(11.9452, 10.2922, 10.0745, 19.6632, 4.2135, 3.7465, 0.114358889272131, 0.0653669981293651,
0.0568296675259594, 0.163542587046242, 0.0569370997973496, 0.0253651116474753,
0.957362700265639, 0.63511200840797, 0.564094173665784, 0.831719084616146,
1.35130176331671, 0.677034876484061),
.Dim = c(6L, 3L),
.Dimnames = list(NULL, c("mean", "sd", "cv")))),
row.names = c(NA, 6L),
class = "data.frame")
查看输出，我们发现它没有很好地表示数据:
test
#> Group.1 Group.2 x.mean x.sd x.cv
#> 1 P1 Compound 1: IgG1 11.94520000 0.11435889 0.95736270
#> 2 P2 Compound 1: IgG1 10.29220000 0.06536700 0.63511201
#> 3 P1 Compound 2: IgG2 10.07450000 0.05682967 0.56409417
#> 4 P2 Compound 2: IgG2 19.66320000 0.16354259 0.83171908
#> 5 P1 Compound 3: IgG3 4.21350000 0.05693710 1.35130176
#> 6 P2 Compound 3: IgG3 3.74650000 0.02536511 0.67703488
write.table(file="test.csv", test, row.names = F, dec=",", sep = ";")
你可以使用:
newdf <- test[1:2] # extract the first two columns
newdf <- cbind(newdf, test$x) # Add `test$x` as 4 new columns
newdf
#> Group.1 Group.2 mean sd cv
#> 1 P1 Compound 1: IgG1 11.9452 0.11435889 0.9573627
#> 2 P2 Compound 1: IgG1 10.2922 0.06536700 0.6351120
#> 3 P1 Compound 2: IgG2 10.0745 0.05682967 0.5640942
#> 4 P2 Compound 2: IgG2 19.6632 0.16354259 0.8317191
#> 5 P1 Compound 3: IgG3 4.2135 0.05693710 1.3513018
#> 6 P2 Compound 3: IgG3 3.7465 0.02536511 0.6770349
write.table(file="test.csv", newdf, row.names = F, dec=",", sep = ";")
由reprex包(v2.0.1)在2018-10-13上创建