Homework 1 Solution
Jeremy King (334007221)
Q1
Part a
canada_rain_df = data.frame(read.table("/Users/Jeremy/downloads/rainfall_60-79.dat"))Part b
nrow(canada_rain_df)## [1] 4826ncol(canada_rain_df)## [1] 27There are 4826 rows and 27 columns.
Part c
canada_rain_df[21,5]## [1] 0The value of canada_rain_df at row 21 column 5 is 0
Part d
canada_rain_df[3,]## V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21
## 3 60 4 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## V22 V23 V24 V25 V26 V27
## 3 0 0 0 0 0 0Part e
names(canada_rain_df) <- c("year","month","day",seq(0,23))
names(canada_rain_df)## [1] "year" "month" "day" "0" "1" "2" "3" "4" "5"
## [10] "6" "7" "8" "9" "10" "11" "12" "13" "14"
## [19] "15" "16" "17" "18" "19" "20" "21" "22" "23"This function creates the labels for the columns of the data frame canada_rain_df. The 0-23 represent the hours of the day.
Part f
daily <- rowSums(canada_rain_df[ , (ncol(canada_rain_df) - 23):ncol(canada_rain_df)], na.rm = TRUE)
hist(daily, main = "Histogram of Daily Rainfall Amount Frequencies", xlab = "Rainfall")
For some reason there are daily values less than 0, so that is why it is messing up the histogram as there is no such thing as negative rainfall.
canada_rain_df[which.min(canada_rain_df$daily),]## [1] year month day 0 1 2 3 4 5 6 7 8
## [13] 9 10 11 12 13 14 15 16 17 18 19 20
## [25] 21 22 23
## <0 rows> (or 0-length row.names)canada_rain_df$daily = NULL
canada_rain_df[canada_rain_df < 0] = NA
daily = rowSums(canada_rain_df[ , (ncol(canada_rain_df) - 23):ncol(canada_rain_df)], na.rm = TRUE)
hist(daily, main = "Histogram of Daily Rainfall Amount Frequencies", xlab = "Rainfall")
Q2
Part a
- vector1 <- c(5, 12, 7, 32) #creates a vector with four elements (5,12,7,32).
- max(vector1) #finds the max element of vector1, which is 32.
- sort(vector1) #sorts the elements of vector1 from least to greatest, making it (5,7,12,32).
- sum(vector1) #adds up all the elements of vector1 together and returns 56.
Part b
vector2 <- c(“5”,7,12)
#creates a vector with 3 elements (“5”,7,12).
vector2[2] + vector2[3]
#doesn’t run as intended as vector2 has non-numeric elements. Because one the elements was initialized as a string, the other numbers got converted to strings.
dataframe3 <- data.frame(z1=“5”,z2=7,z3=12)
#creates a data.frame with elements of (“5”,7,12)
dataframe3[1,2] + dataframe3[1,3]
#works as intended as the data.frame stores elements in their original form as they were when they were initialized. Thus, 7 and 12 are still numeric values.
list4 <- list(z1=“6”, z2=42, z3=“49”, z4=126)
#creates a list with elements 4 elements (“6”,42,“49”,126).
list4[[2]]+list4[[4]]
#works as intended as it calls the numeric value directly from list and returns 168
list4[2]+list4[4]
#doesn’t work as intended as list4[2] = z2, which is not a numerical value and thus cannot add them together.
Q3
Part a
seq(from = 1, to = 10000, by = 369)## [1] 1 370 739 1108 1477 1846 2215 2584 2953 3322 3691 4060 4429 4798 5167
## [16] 5536 5905 6274 6643 7012 7381 7750 8119 8488 8857 9226 9595 9964seq(from = 1, to = 10000, length.out = 50)## [1] 1.0000 205.0612 409.1224 613.1837 817.2449 1021.3061
## [7] 1225.3673 1429.4286 1633.4898 1837.5510 2041.6122 2245.6735
## [13] 2449.7347 2653.7959 2857.8571 3061.9184 3265.9796 3470.0408
## [19] 3674.1020 3878.1633 4082.2245 4286.2857 4490.3469 4694.4082
## [25] 4898.4694 5102.5306 5306.5918 5510.6531 5714.7143 5918.7755
## [31] 6122.8367 6326.8980 6530.9592 6735.0204 6939.0816 7143.1429
## [37] 7347.2041 7551.2653 7755.3265 7959.3878 8163.4490 8367.5102
## [43] 8571.5714 8775.6327 8979.6939 9183.7551 9387.8163 9591.8776
## [49] 9795.9388 10000.0000Part b
rep(1:3, times = 3)## [1] 1 2 3 1 2 3 1 2 3rep(1:3, each = 3)## [1] 1 1 1 2 2 2 3 3 3As shown in the code, the first line goes from 1-3 3 times (1 2 3 1…) while the second line prints each element of 1-3 3 times (1 1 1 2…).