我得到以下结果，我试图在管道命令中应用一个函数。
我正在使用的代码给我以下错误：
sample_rmse_tbl <- dataset %>%
mutate(rmse = map_dbl(predict, calc_rmse)) %>%
select(id, rmse)
哪个出现以下错误：
Error in mutate_impl(.data, dots) :
Evaluation error: Result 1 is not a length 1 atomic vector.
数据使用太阳黑子数据，下面是我的代码（我遇到的错误是代码的最后一行）：
我已经仔细阅读了本教程，直到这行代码为止，一切对我都有效。
---代码是本教程的精简版：https://www.business-science.io/timeseries-analysis/2018/04/18/keras-lstm-sunspots-time-series-prediction.html
# Core Tidyverse
library(tidyverse)
library(glue)
library(forcats)
# Time Series
library(timetk)
library(tidyquant)
library(tibbletime)
# Visualization
library(cowplot)
# Preprocessing
library(recipes)
# Sampling / Accuracy
library(rsample)
library(yardstick)
# Modeling
library(keras)
sun_spots <- datasets::sunspot.month %>%
tk_tbl() %>%
mutate(index = as_date(index)) %>%
as_tbl_time(index = index)
sun_spots
############################################
periods_train <- 12 * 50
periods_test <- 12 * 10
skip_span <- 12 * 20
rolling_origin_resamples <- rolling_origin(
sun_spots,
initial = periods_train,
assess = periods_test,
cumulative = FALSE,
skip = skip_span
)
rolling_origin_resamples
############################################
calc_rmse <- function(prediction_tbl) {
rmse_calculation <- function(data) {
data %>%
spread(key = key, value = value) %>%
select(-index) %>%
filter(!is.na(predict)) %>%
rename(
truth = actual,
estimate = predict
) %>%
rmse(truth, estimate)
}
safe_rmse <- possibly(rmse_calculation, otherwise = NA)
safe_rmse(prediction_tbl)
}
#############################################
predict_keras_lstm <- function(split, epochs = 300, ...) {
lstm_prediction <- function(split, epochs, ...) {
# 5.1.2 Data Setup
df_trn <- training(split)
df_tst <- testing(split)
df <- bind_rows(
df_trn %>% add_column(key = "training"),
df_tst %>% add_column(key = "testing")
) %>%
as_tbl_time(index = index)
# 5.1.3 Preprocessing
rec_obj <- recipe(value ~ ., df) %>%
step_sqrt(value) %>%
step_center(value) %>%
step_scale(value) %>%
prep()
df_processed_tbl <- bake(rec_obj, df)
center_history <- rec_obj$steps[[2]]$means["value"]
scale_history <- rec_obj$steps[[3]]$sds["value"]
# 5.1.4 LSTM Plan
lag_setting <- 120 # = nrow(df_tst)
batch_size <- 40
train_length <- 440
tsteps <- 1
epochs <- epochs
# 5.1.5 Train/Test Setup
lag_train_tbl <- df_processed_tbl %>%
mutate(value_lag = lag(value, n = lag_setting)) %>%
filter(!is.na(value_lag)) %>%
filter(key == "training") %>%
tail(train_length)
x_train_vec <- lag_train_tbl$value_lag
x_train_arr <- array(data = x_train_vec, dim = c(length(x_train_vec), 1, 1))
y_train_vec <- lag_train_tbl$value
y_train_arr <- array(data = y_train_vec, dim = c(length(y_train_vec), 1))
lag_test_tbl <- df_processed_tbl %>%
mutate(
value_lag = lag(value, n = lag_setting)
) %>%
filter(!is.na(value_lag)) %>%
filter(key == "testing")
x_test_vec <- lag_test_tbl$value_lag
x_test_arr <- array(data = x_test_vec, dim = c(length(x_test_vec), 1, 1))
y_test_vec <- lag_test_tbl$value
y_test_arr <- array(data = y_test_vec, dim = c(length(y_test_vec), 1))
# 5.1.6 LSTM Model
model <- keras_model_sequential()
model %>%
layer_lstm(units = 50,
input_shape = c(tsteps, 1),
batch_size = batch_size,
return_sequences = TRUE,
stateful = TRUE) %>%
layer_lstm(units = 50,
return_sequences = FALSE,
stateful = TRUE) %>%
layer_dense(units = 1)
model %>%
compile(loss = 'mae', optimizer = 'adam')
# 5.1.7 Fitting LSTM
for (i in 1:epochs) {
model %>% fit(x = x_train_arr,
y = y_train_arr,
batch_size = batch_size,
epochs = 1,
verbose = 1,
shuffle = FALSE)
model %>% reset_states()
cat("Epoch: ", i)
}
# 5.1.8 Predict and Return Tidy Data
# Make Predictions
pred_out <- model %>%
predict(x_test_arr, batch_size = batch_size) %>%
.[,1]
# Retransform values
pred_tbl <- tibble(
index = lag_test_tbl$index,
value = (pred_out * scale_history + center_history)^2
)
# Combine actual data with predictions
tbl_1 <- df_trn %>%
add_column(key = "actual")
tbl_2 <- df_tst %>%
add_column(key = "actual")
tbl_3 <- pred_tbl %>%
add_column(key = "predict")
# Create time_bind_rows() to solve dplyr issue
time_bind_rows <- function(data_1, data_2, index) {
index_expr <- enquo(index)
bind_rows(data_1, data_2) %>%
as_tbl_time(index = !! index_expr)
}
ret <- list(tbl_1, tbl_2, tbl_3) %>%
reduce(time_bind_rows, index = index) %>%
arrange(key, index) %>%
mutate(key = as_factor(key))
return(ret)
}
safe_lstm <- possibly(lstm_prediction, otherwise = NA)
safe_lstm(split, epochs, ...)
}
#################################################
sample_predictions_lstm_tbl <- rolling_origin_resamples %>%
mutate(predict = map(splits, predict_keras_lstm, epochs = 10))
sample_predictions_lstm_tbl
sample_predictions_lstm_tbl$predict
map_dbl(sample_predictions_lstm_tbl$predict, calc_rmse)
sample_rmse_tbl <- sample_predictions_lstm_tbl %>%
mutate(rmse = map_dbl(predict, calc_rmse)) %>%
select(id, rmse)
EDIT1：
[[11]]
# A time tibble: 840 x 3
# Index: index
index value key
<date> <dbl> <fct>
1 1949-11-01 144. actual
2 1949-12-01 118. actual
3 1950-01-01 102. actual
4 1950-02-01 94.8 actual
5 1950-03-01 110. actual
6 1950-04-01 113. actual
7 1950-05-01 106. actual
8 1950-06-01 83.6 actual
9 1950-07-01 91 actual
10 1950-08-01 85.2 actual
# ... with 830 more rows
EDIT2：
我想出了一种“解决方法”，但与本文的结果有所不同。
temp <- NULL
sample_rmse_tbl <- NULL
for(i in 1:length(sample_predictions_lstm_tbl$predict)){
temp <- calc_rmse(sample_predictions_lstm_tbl$predict[[i]])
sample_rmse_tbl[[i]] <- temp
}
sample_rmse_tbl <- do.call(rbind.data.frame, sample_rmse_tbl)
sample_rmse_tbl %>%
setNames(., c("metric", "estimator", "rmse")) %>%
mutate(id = row_number()) %>%
select(id, rmse)
答案 0 :(得分：1)
问题在于函数rmse()返回的是列表，而不是单个double值。您需要使用.$.estimate从此列表中选择估计值。但是，我必须删除possibly()调用才能使解决方案正常工作。
因此，新功能calc_rmse()如下所示。
calc_rmse <- function(prediction_tbl) {
rmse_calculation <- function(data) {
data %>%
spread(key = key, value = value) %>%
select(-index) %>%
filter(!is.na(predict)) %>%
rename(
truth = actual,
estimate = predict
) %>%
rmse(truth, estimate) %>% .$.estimate
}
rmse_calculation(prediction_tbl)
}