Great job providing code and context! Let's break down your **error** and how to address it, and then I'll give some guidance for your **two questions**. --- ## 1. The Error Explained Your error message is: ``` Error in plot.default(...) : formal argument "type" matched by multiple actual arguments ``` This usually happens when you give the `plot()` function an argument (like `type`) that conflicts with how R interprets your data input. Here's your command: ```r plot(BikeDemand, type = "l", col = 1, ylim = c(0, 35), main = "Types of moving averages", sub = "Oct 3 2016--Oct 23 2016", xlab = "Hours", ylab = "Bike Usage/Demand") ``` **Problem:** `BikeDemand` is a data frame (not a single vector), and when you call `plot()` on a data frame, R tries to guess what you want. It might try to plot a matrix of scatterplots or, in some cases, treat it as a formula. The `type` argument is *not* accepted in all these call signatures, causing the error. --- ## 2. Solution: How to Plot Time Series for Each Station ### **Option 1: Plot each station individually** Suppose your columns are: `Time, Station1, Station2, ..., Station6`. ```r # First column is time, columns 2-7 are stations time <- BikeDemand[[1]] # Plot Station 1 plot(time, BikeDemand[[2]], type="l", col=2, ylim=c(0, 35), main="Bike Demand at Station 1", xlab="Hour", ylab="Bike Usage/Demand") ``` ### **Option 2: Plot all stations in one plot** ```r matplot(time, as.matrix(BikeDemand[,-1]), type="l", lty=1, col=1:6, ylim=c(0, 35), main="Bike Demand at 6 Stations", xlab="Hour", ylab="Bike Usage/Demand") legend("topright", legend=names(BikeDemand)[-1], col=1:6, lty=1) ``` --- ## 3. Answering Your Two Questions ### **1. Predict next 3 hours of bike station demands** You could use a simple moving average (your `movavg` function) or a more sophisticated time series model (like ARIMA from the `forecast` package). #### **Using your moving average function:** ```r n <- 3 # or whatever window you want predictions <- sapply(BikeDemand[,-1], function(x) { y <- movavg(x, n, type = "s") # Simple way to predict: repeat last moving average value for next 3 hours rep(y[length(y)], 3) }) # predictions is now a 6x3 matrix: 3 predicted values for each station ``` #### **Discussion:** With the moving average, you're assuming the immediate past trend continues. For more sophisticated modeling, use `auto.arima()` from the `forecast` package. ### **2. Aggregate hourly data into daily data, predict the next 3 days** #### **Aggregate:** ```r library(dplyr) BikeDemand$Day <- as.integer((seq_len(nrow(BikeDemand)) - 1) / 24) + 1 daily_demand <- BikeDemand %>% group_by(Day) %>% summarise(across(-c(Time, Day), sum)) # Sum demand per day per station ``` #### **Predict next 3 days (using moving average):** ```r daily_pred <- sapply(daily_demand[,-1], function(x) { y <- movavg(x, n=3, type="s") rep(y[length(y)], 3) }) ``` #### **Discussion:** Aggregating smooths out hourly fluctuations. If trends are stable, predictions may be decent; if there's seasonality or strong trends, moving average may not predict well. --- ## **Summary of Your Issue** - **Plot error**: Caused by using `plot()` directly on a data frame with `type="l"`. - **Solution**: Use `matplot()` for multiple columns, or plot individual columns. - **Prediction**: Use your moving average function for a quick forecast, or use more advanced time series models for better accuracy. - **Aggregation**: Use `dplyr` to sum hourly to daily data. --- **If you need more detail on forecasting (e.g., using `forecast::auto.arima()`), just ask!**