更改数据框中的一些数值 [英] Change some numerical values in data frame
问题描述
我有一个数据框,其中两个变量(Lat 和 Lon)的值不正确.数据框中不正确的值被列为 999.00,正确的值应该分别是 42.68 和 -72.47.
我想要一种使用 dplyr 替换这些值的简单方法,但我的尝试(见下文)未成功(下面提供了错误).
df$Lat2 <- recode(df$Lat, "999.00"="42.68", .default=x)
<块引用>
lapply(x, f) 中的错误:找不到对象x"
df <- df %>%变异(Lat2 = if_else(Lat == 999.00, 42.68, NULL, NULL))
<块引用>
mutate_impl(.data, dots) 中的错误:评估错误:未使用的参数 (recvLat = 999).
df <- df %>%变异(Lat2 = ifelse(Lat == 999.00, 42.68, NULL))
<块引用>
mutate_impl(.data, dots) 中的错误:评估错误:替换长度为零.另外: 警告信息:在 rep(no, length.out = length(ans)) 中:'x' 为 NULL,因此结果将为 NULL
df <- df %>%变异(Lat2 = case_when(Lat == 999.00 ~ 42.68, TRUE ~ NULL))
<块引用>
mutate_impl(.data, dots) 中的错误:评估错误:下标越界.
对于后三个尝试,如果数字在引号中(即999.00"和42.68"),我会得到相同的错误
我们可以将 NULL
放入 list
df %>%变异(Lat2 = ifelse(recvLat == 999.00, 42.68, list(NULL)))# recvLat Lat2#1 999.0 42.68#2 1.5 空#3 2.5 空
代替NULL
,它可以是NA
df %>%变异(Lat2 = ifelse(recvLat == 999.00, 42.68, NA_real_))# recvLat Lat2#1 999.0 42.68#2 1.5 北美#3 2.5 北美
如果我们想做相反的事情,只需使用 !=
df %>%变异(Lat2 = ifelse(recvLat != 999.00, 42.68, NA_real_))
根据 OP 的评论,
df %>%变异(Lat2 = ifelse(recvLat == 999.00, 42.68, recvLat))
在base R
中,我们可以通过创建索引来做到这一点
i1 <- df$recvLat == 999df$recLat[i1] <- 42.68
注意:两种解决方案都有效.
数据
df <- data.frame(recvLat = c(999, 1.5, 2.5))
I have a dataframe with incorrect values for two variables (Lat and Lon). The incorrect values in the dataframe are listed as 999.00, and the correct values should be 42.68 and -72.47, respectively.
I would like an easy way to replace these values using dplyr, but my attempts (see below) have been unsuccessful (errors provided below).
df$Lat2 <- recode(df$Lat, "999.00"="42.68", .default=x)
Error in lapply(x, f) : object 'x' not found
df <- df %>%
mutate(Lat2 = if_else(Lat == 999.00, 42.68, NULL, NULL))
Error in mutate_impl(.data, dots) : Evaluation error: unused argument (recvLat = 999).
df <- df %>%
mutate(Lat2 = ifelse(Lat == 999.00, 42.68, NULL))
Error in mutate_impl(.data, dots) : Evaluation error: replacement has length zero. In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL
df <- df %>%
mutate(Lat2 = case_when(Lat == 999.00 ~ 42.68, TRUE ~ NULL))
Error in mutate_impl(.data, dots) : Evaluation error: subscript out of bounds.
For the latter three attempts, I get the same error if the number are in quotes (i.e. "999.00" and "42.68")
We can place NULL
in a list
df %>%
mutate(Lat2 = ifelse(recvLat == 999.00, 42.68, list(NULL)))
# recvLat Lat2
#1 999.0 42.68
#2 1.5 NULL
#3 2.5 NULL
Instead of NULL
, it can be NA
df %>%
mutate(Lat2 = ifelse(recvLat == 999.00, 42.68, NA_real_))
# recvLat Lat2
#1 999.0 42.68
#2 1.5 NA
#3 2.5 NA
If we want to do the opposite, just use !=
df %>%
mutate(Lat2 = ifelse(recvLat != 999.00, 42.68, NA_real_))
Based on the OP's comment,
df %>%
mutate(Lat2 = ifelse(recvLat == 999.00, 42.68, recvLat))
In base R
, we can do this by creating an index
i1 <- df$recvLat == 999
df$recLat[i1] <- 42.68
NOTE: Both the solutions work.
data
df <- data.frame(recvLat = c(999, 1.5, 2.5))
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