我的CSV看起来像这样:
1991 1992 1993 1991 1992 1993
VariableA VariableB VariableC VariableC VariableC VariableD VariableD VariableD
lm mt 1 3 4 2 3 5
所以我想创建一个名为year的变量并执行以下操作:
VariableA VariableB Year VariableC VariableD
lm mt 1991 1 2
lm mt 1992 3 3
lm mt 1993 4 5
我主要使用Pandas,我正在学习,但我不知道正确读取数据,然后进行操作。如果有人在R中提出解决方案,那也会很好。
EDITION:
我的真实数据框架是从1991年到2013年的多年,并且有更多变量具有重复。我使用Ananda Mahto建议的包splitstackshape
尝试了代码inr R,但是我收到了一条错误消息。那么,我的错误是什么?
mydf <- read.csv("DatosCOMPUSTATfinal.csv", skip = 3, check.names = FALSE)
nombres <- names(mydf)[-c(1,2,3)]
nombres <- unique(nombres)
> nombres
[1] "Employees" "Market Value-daily"
[3] "Market to book - daily" "Total return"
[5] "Total assets" "total stockholders' equity"
[7] "Sales" "EBITDA"
[9] "EBIT" "Pretax income"
[11] "Income (loss)"
> names(mydf[c(1,2,3)])
[1] "Company name" "employer identification"
[3] "CUSIP"
names(mydf)[-c(1,2,3)] <- paste(names(mydf)[-c(1,2,3)],
c(1991:2013), sep = "_")
nv <- merged.stack(mydf, id.vars = names(mydf[c(1,2,3)]) , var.stubs = nombres , sep = "_" )
然后,我收到错误消息:
Error in if (ncol(x) == 1L) { : argument is of length zero
第2版:
我用reshape函数尝试了这段代码,但收到了“内存耗尽”的消息。我不知道为什么,因为数据框只是改变它的方向,它的大小小于15 mb。为什么会发生这种情况?我该如何处理?
newmydf <- reshape(mydf, direction = "long", idvar = 1:3, varying = 4:ncol(mydf), sep = "_")
Error: memory exhausted (limit reached?)
答案 0 :(得分:7)
在R中,一种方法可能是读取跳过第一行的csv,将其作为变量名称的一部分添加回来,然后使用reshape
获取所需的输出。
尝试以下内容:
mydf <- read.csv("yourfile.csv", skip = 1, check.names = FALSE)
names(mydf)[-c(1, 2)] <- paste(names(mydf)[-c(1, 2)],
c(1991, 1992, 1993), sep = "_")
reshape(mydf, direction = "long", idvar = 1:2,
varying = 3:ncol(mydf), sep = "_")
# VariableA VariableB time VariableC VariableD
# lm.mt.1991 lm mt 1991 1 2
# lm.mt.1992 lm mt 1992 3 3
# lm.mt.1993 lm mt 1993 4 5
重命名步骤后,如果reshape()
对您来说太慢,请尝试使用我的“splitstackshape”软件包中的merged.stack
:
library(splitstackshape)
merged.stack(mydf, var.stubs = c("VariableC", "VariableD"), sep = "_")
# VariableA VariableB .time_1 VariableC VariableD
# 1: lm mt 1991 1 2
# 2: lm mt 1992 3 3
# 3: lm mt 1993 4 5
答案 1 :(得分:1)
R
中的另一种方法是在使用dplyr/tidyr
读取数据集之后使用read.csv
(对于大数据集会更快),如@Ananda Mahto的帖子中所述
library(dplyr)
library(tidyr)
mydf %>%
gather(Var, Val, matches("[0-9]+$")) %>%
separate(Var, c("Var", "Year")) %>%
spread(Var, Val)
# VariableA VariableB Year VariableC VariableD
#1 lm mt 1991 1 2
#2 lm mt 1992 3 3
#3 lm mt 1993 4 5
mydf <- structure(list(VariableA = structure(1L, .Label = "lm", class = "factor"),
VariableB = structure(1L, .Label = "mt", class = "factor"),
VariableC_1991 = 1L, VariableC_1992 = 3L, VariableC_1993 = 4L,
VariableD_1991 = 2L, VariableD_1992 = 3L, VariableD_1993 = 5L), .Names = c("VariableA",
"VariableB", "VariableC_1991", "VariableC_1992", "VariableC_1993",
"VariableD_1991", "VariableD_1992", "VariableD_1993"), class = "data.frame", row.names = c(NA,
-1L))