如何在剪切

Question

我希望根据break参数中的条件语句cut我的数据，但它会抛出此错误：

  

矩阵错误（unlist（value，recursive = FALSE，use.names = FALSE），   nrow = nr，：dimnames的长度＆＃39; [2]不等于数组范围

是否可以这样在cut内使用条件语句？

示例数据

df <- structure(list(fyear = c(1970, 1970, 1970, 1970, 1970, 1970, 
1970, 1970, 1970, 1970, 1970, 1970, 1970, 1970, 1970, 1970, 1970, 
1970, 1970, 1970), BEME = c(0.39713747645951, 0.548988782444936, 
0.537154930871343, 1.89357008340059, 1.66945262543448, 0.969181836638018, 
1.09989952916609, 0.858308443214104, 0.292175536881419, 0.684685677549708, 
0.338422675433708, 3.02671555788371, 0.422643864469658, 0.805317430736738, 
0.529954031556715, 0.617716486520065, 0.911576593365635, 0.4131850675139, 
1.16211278792693, 2.13177678851802), exchg = c(11L, 11L, 11L, 
11L, 11L, 11L, 11L, 11L, 12L, 12L, 12L, 11L, 11L, 12L, 11L, 12L, 
19L, 11L, 11L, 11L)), .Names = c("fyear", "BEME", "exchg"), class = c("tbl_df", 
"data.frame"), row.names = c(NA, -20L))

Cut功能

cut(df$BEME, breaks = quantile(df[df$exchg == 11, 2], c(0,0.3,0.7,1)), labels = FALSE)

Answer 1

在

cut(df$BEME, breaks = quantile(df[df$exchg == 11, 2]$BEME, c(0,0.3,0.7,1)), labels = FALSE)

变化

df[df$exchg == 11, 2]

到

df[df$exchg == 11, 2]$BEME

第一项返回data.frame第二个向量（这是你想要的）。