Question

这与帖子＆＃34; Replace all values in a column of one dataframe using values in a row of another dataframe (matching by row name and column name)＆＃34;

非常相似

除了我的替换要求之外（＆＃34; 1＆＃34;）被其所在列的名称替换。

示例设置

df1<-data.frame(replicate(5,sample(0:1,5,rep=TRUE)))
row.names(df1)<-c("hootsuite","foodtank","FarmsNews","agchat","TysonFoods")
names(df1)<-c("food","agvocate","editor","gmo","ag")

           food agvocate editor gmo ag
hootsuite     1        1      0   0  1
foodtank      1        1      0   0  1
FarmsNews     1        0      1   0  1
agchat        0        0      0   0  1
TysonFoods    1        0      1   1  0

会变成

                food    agvocate    editor gmo      ag
hootsuite       food    agvocate    0       0       ag
foodtank        food    agvocate    0       0       ag
FarmsNews       food    0           editor  0       ag
agchat          0       0           0       0       ag
TysonFoods      food    0           editor  gmo     0

similar post

中的解决方案

df1*df2[,1][col(df1)] 
or  
sweep(df1, 2, df2[,1], "*")

（使用下面定义的df2）

df2<-c("food","agvocate","editor","gmo","ag")
df2<-as.matrix(df2)
row.names(df2)<-c("food","agvocate","editor","gmo","ag")

给出错误＆＃34; FUN中的错误（左，右）：二元运算符＆＃34; 的非数字参数，意味着矩阵乘法实际上不适用于字符;）

所以我应该采取的方法是什么？

Answer 1

我们可以使用OnClickListener到Map相应列，其值为1，列名为

replace

或使用逻辑矩阵进行子集化和分配

df1[] <- Map(function(x, y) replace(x, x==1, y), df1, names(df1))
df1
#           food agvocate editor gmo ag
#hootsuite  food agvocate      0   0 ag
#foodtank   food agvocate      0   0 ag
#FarmsNews  food        0 editor   0 ag
#agchat        0        0      0   0 ag
#TysonFoods food        0 editor gmo  0

Answer 2

使用purrr包的解决方案。 map2_dfc可以使用函数循环遍历两个列表或向量（在本例中为df1和names(df1)）。它还可以将输出绑定为数据帧。

set.seed(1)

df1<-data.frame(replicate(5,sample(0:1,5,rep=TRUE)))
row.names(df1)<-c("hootsuite","foodtank","FarmsNews","agchat","TysonFoods")
names(df1)<-c("food","agvocate","editor","gmo","ag")

library(purrr)

df2 <- map2_dfc(df1, names(df1), function(x, y){
  x[x %in% 1] <- y
  return(x)
})
df2
# # A tibble: 5 x 5
#    food agvocate editor   gmo    ag
#   <chr>    <chr>  <chr> <chr> <chr>
# 1     0 agvocate      0     0    ag
# 2     0 agvocate      0   gmo     0
# 3  food agvocate editor   gmo    ag
# 4  food agvocate      0     0     0
# 5     0        0 editor   gmo     0

使用另一个数据帧的行中的值替换一个数据帧的列中的所有值（按行名称和列名称匹配），替换为字符

2 个答案: