Take the mean of a columns in csv and create a new row with the information
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I am pretty new to R but liked all of the help that was available. The problem I am encountering, is that there have been so many updates since a lot of videos and help sites have been made that the code is no longer valid.
I have a csv of immunization percentages and I want to find the mean of each column of data. If there is a way to add this value as the last row of data that would be ideal but not necessary. I have tried several different types of mean functions but I continue to get error messages.
I have tried the following codes. My file is called Measles1 and the columns are Y followed by the year. I took put the error codes specific to the lines but wanted to show what I have tried for reference. Any help would be greatly appreciated.
> colMeans(Measles1$Y2017)
> colMeans(Measles1)
> mean(Measles1$Y2017)
> mean(Measles1$Y2017, na.rm = TRUE)
> colMeans(Measles1$Y2017, na.rm = TRUE)
> Means <- colMeans(as.numeric(as.character(Measles1)))
> results.mean <- mean(Measles1)
> results.mean <- mean(Measles1,na.rm = TRUE)
> mean(Measles1[2:39])
I am sure that I am just missing something very simple. Thank you for your help.
r row rstudio add mean
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
add a comment |
I am pretty new to R but liked all of the help that was available. The problem I am encountering, is that there have been so many updates since a lot of videos and help sites have been made that the code is no longer valid.
I have a csv of immunization percentages and I want to find the mean of each column of data. If there is a way to add this value as the last row of data that would be ideal but not necessary. I have tried several different types of mean functions but I continue to get error messages.
I have tried the following codes. My file is called Measles1 and the columns are Y followed by the year. I took put the error codes specific to the lines but wanted to show what I have tried for reference. Any help would be greatly appreciated.
> colMeans(Measles1$Y2017)
> colMeans(Measles1)
> mean(Measles1$Y2017)
> mean(Measles1$Y2017, na.rm = TRUE)
> colMeans(Measles1$Y2017, na.rm = TRUE)
> Means <- colMeans(as.numeric(as.character(Measles1)))
> results.mean <- mean(Measles1)
> results.mean <- mean(Measles1,na.rm = TRUE)
> mean(Measles1[2:39])
I am sure that I am just missing something very simple. Thank you for your help.
r row rstudio add mean
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
Trynew_Measles <- rbind(Measles1, colMeans(Measles1))
– Ronak Shah
Nov 29 '18 at 3:46
add a comment |
I am pretty new to R but liked all of the help that was available. The problem I am encountering, is that there have been so many updates since a lot of videos and help sites have been made that the code is no longer valid.
I have a csv of immunization percentages and I want to find the mean of each column of data. If there is a way to add this value as the last row of data that would be ideal but not necessary. I have tried several different types of mean functions but I continue to get error messages.
I have tried the following codes. My file is called Measles1 and the columns are Y followed by the year. I took put the error codes specific to the lines but wanted to show what I have tried for reference. Any help would be greatly appreciated.
> colMeans(Measles1$Y2017)
> colMeans(Measles1)
> mean(Measles1$Y2017)
> mean(Measles1$Y2017, na.rm = TRUE)
> colMeans(Measles1$Y2017, na.rm = TRUE)
> Means <- colMeans(as.numeric(as.character(Measles1)))
> results.mean <- mean(Measles1)
> results.mean <- mean(Measles1,na.rm = TRUE)
> mean(Measles1[2:39])
I am sure that I am just missing something very simple. Thank you for your help.
r row rstudio add mean
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
I am pretty new to R but liked all of the help that was available. The problem I am encountering, is that there have been so many updates since a lot of videos and help sites have been made that the code is no longer valid.
I have a csv of immunization percentages and I want to find the mean of each column of data. If there is a way to add this value as the last row of data that would be ideal but not necessary. I have tried several different types of mean functions but I continue to get error messages.
I have tried the following codes. My file is called Measles1 and the columns are Y followed by the year. I took put the error codes specific to the lines but wanted to show what I have tried for reference. Any help would be greatly appreciated.
> colMeans(Measles1$Y2017)
> colMeans(Measles1)
> mean(Measles1$Y2017)
> mean(Measles1$Y2017, na.rm = TRUE)
> colMeans(Measles1$Y2017, na.rm = TRUE)
> Means <- colMeans(as.numeric(as.character(Measles1)))
> results.mean <- mean(Measles1)
> results.mean <- mean(Measles1,na.rm = TRUE)
> mean(Measles1[2:39])
I am sure that I am just missing something very simple. Thank you for your help.
r row rstudio add mean
r row rstudio add mean
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
asked Nov 29 '18 at 3:43
SStephanieSStephanie
91
91
Trynew_Measles <- rbind(Measles1, colMeans(Measles1))
– Ronak Shah
Nov 29 '18 at 3:46
add a comment |
Trynew_Measles <- rbind(Measles1, colMeans(Measles1))
– Ronak Shah
Nov 29 '18 at 3:46
Try
new_Measles <- rbind(Measles1, colMeans(Measles1))– Ronak Shah
Nov 29 '18 at 3:46
Try
new_Measles <- rbind(Measles1, colMeans(Measles1))– Ronak Shah
Nov 29 '18 at 3:46
add a comment |
2 Answers
2
active
oldest
votes
It would give us a better idea if you could provide a representative sample. You need to make sure that all columns are numeric in order to compute their means at once. One way to check this would be str(your_data_frame).
Using the built-in mtcars dataset:
# na.rm argument is optional depending on your data
mtcars[nrow(mtcars) + 1, ] <- colMeans(mtcars, na.rm = T)
@Ronak Shah's recommendation works well, too:
mtcars <- rbind(mtcars, colMeans(mtcars, na.rm = T))
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
add a comment |
This code will provide you option of finding grand total or mean by column values
d1 <- data_frame(
name = c("jim", "john", "jim", "john"),
`2012` = c(57, 58, 47, 57),
`2013` = c(14, 3, 3, 90))
library(tidyverse)
d1 <-bind_rows(d1,
d1 %>%
group_by(name) %>%
summarise_all(funs(mean)) %>%
mutate(name = paste0(name, '_total')))
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It would give us a better idea if you could provide a representative sample. You need to make sure that all columns are numeric in order to compute their means at once. One way to check this would be str(your_data_frame).
Using the built-in mtcars dataset:
# na.rm argument is optional depending on your data
mtcars[nrow(mtcars) + 1, ] <- colMeans(mtcars, na.rm = T)
@Ronak Shah's recommendation works well, too:
mtcars <- rbind(mtcars, colMeans(mtcars, na.rm = T))
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
add a comment |
It would give us a better idea if you could provide a representative sample. You need to make sure that all columns are numeric in order to compute their means at once. One way to check this would be str(your_data_frame).
Using the built-in mtcars dataset:
# na.rm argument is optional depending on your data
mtcars[nrow(mtcars) + 1, ] <- colMeans(mtcars, na.rm = T)
@Ronak Shah's recommendation works well, too:
mtcars <- rbind(mtcars, colMeans(mtcars, na.rm = T))
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
add a comment |
It would give us a better idea if you could provide a representative sample. You need to make sure that all columns are numeric in order to compute their means at once. One way to check this would be str(your_data_frame).
Using the built-in mtcars dataset:
# na.rm argument is optional depending on your data
mtcars[nrow(mtcars) + 1, ] <- colMeans(mtcars, na.rm = T)
@Ronak Shah's recommendation works well, too:
mtcars <- rbind(mtcars, colMeans(mtcars, na.rm = T))
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
It would give us a better idea if you could provide a representative sample. You need to make sure that all columns are numeric in order to compute their means at once. One way to check this would be str(your_data_frame).
Using the built-in mtcars dataset:
# na.rm argument is optional depending on your data
mtcars[nrow(mtcars) + 1, ] <- colMeans(mtcars, na.rm = T)
@Ronak Shah's recommendation works well, too:
mtcars <- rbind(mtcars, colMeans(mtcars, na.rm = T))
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
answered Nov 29 '18 at 4:02
Ozan147Ozan147
2,0821519
2,0821519
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
add a comment |
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
you were correct the data was not being stored as numeric. I fixed that and tried to use the code and realized the first column are the country names instead of being the titles of the rows. Once I fixed all of this the code worked perfect.
– SStephanie
Nov 30 '18 at 4:08
add a comment |
This code will provide you option of finding grand total or mean by column values
d1 <- data_frame(
name = c("jim", "john", "jim", "john"),
`2012` = c(57, 58, 47, 57),
`2013` = c(14, 3, 3, 90))
library(tidyverse)
d1 <-bind_rows(d1,
d1 %>%
group_by(name) %>%
summarise_all(funs(mean)) %>%
mutate(name = paste0(name, '_total')))
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
add a comment |
This code will provide you option of finding grand total or mean by column values
d1 <- data_frame(
name = c("jim", "john", "jim", "john"),
`2012` = c(57, 58, 47, 57),
`2013` = c(14, 3, 3, 90))
library(tidyverse)
d1 <-bind_rows(d1,
d1 %>%
group_by(name) %>%
summarise_all(funs(mean)) %>%
mutate(name = paste0(name, '_total')))
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
add a comment |
This code will provide you option of finding grand total or mean by column values
d1 <- data_frame(
name = c("jim", "john", "jim", "john"),
`2012` = c(57, 58, 47, 57),
`2013` = c(14, 3, 3, 90))
library(tidyverse)
d1 <-bind_rows(d1,
d1 %>%
group_by(name) %>%
summarise_all(funs(mean)) %>%
mutate(name = paste0(name, '_total')))
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
This code will provide you option of finding grand total or mean by column values
d1 <- data_frame(
name = c("jim", "john", "jim", "john"),
`2012` = c(57, 58, 47, 57),
`2013` = c(14, 3, 3, 90))
library(tidyverse)
d1 <-bind_rows(d1,
d1 %>%
group_by(name) %>%
summarise_all(funs(mean)) %>%
mutate(name = paste0(name, '_total')))
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
answered Nov 29 '18 at 4:14
HunaidkhanHunaidkhan
1,024516
1,024516
add a comment |
add a comment |
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new_Measles <- rbind(Measles1, colMeans(Measles1))– Ronak Shah
Nov 29 '18 at 3:46