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How to create a new column in a data frame depending on multiple criteria from multiple columns from the same data frame
How to sort a dataframe by multiple column(s)How to join (merge) data frames (inner, outer, left, right)Split data frame string column into multiple columnsDrop data frame columns by nameHow to drop columns by name in a data frameChanging column names of a data frameExtracting specific columns from a data frameConvert object list to obtain rownames RBaum-Welch algorithm showing Log-likelihood: NaN BIC criterium: NaN AIC criterium: NaN
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I have a data frame df1 with four variables. One refers to sunlight, the second one refers to the moon-phase light (light due to the moon's phase), the third one to the moon-position light (light from the moon depending on if it is in the sky or not) and the fourth refers to the clarity of the sky (opposite to cloudiness).
I call them SL, MPhL, MPL and SC respectively. I want to create a new column referred to "global light" that during the day depends only on SL and during the night depends on the other three columns ("MPhL", "MPL" and "SC"). What I want is that at night (when SL == 0), the light in a specific area is equal to the product of the columns "MPhL", "MPL" and "SC". If any of them is 0, then, the light at night would be 0 also.
Since I work with a matrix of hundreds of thousands of rows, what would be the best way to do it? As an example of what I have:
SL<- c(0.82,0.00,0.24,0.00,0.98,0.24,0.00,0.00)
MPhL<- c(0.95,0.85,0.65,0.35,0.15,0.00,0.87,0.74)
MPL<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
SC<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
df<-data.frame(SL,MPhL,MPL,SC)
df
SL MPhL MPL SC
1 0.82 0.95 0.00 0.00
2 0.00 0.85 0.50 0.50
3 0.24 0.65 0.10 0.10
4 0.00 0.35 0.89 0.89
5 0.98 0.15 0.33 0.33
6 0.24 0.00 0.58 0.58
7 0.00 0.87 0.00 0.00
8 0.00 0.74 0.46 0.46
What I would like to get is this:
df
SL MPhL MPL SC GL
1 0.82 0.95 0.00 0.00 0.82 # When "SL">0, GL= SL
2 0.00 0.85 0.50 0.50 0.21 # When "SL" is 0, GL = MPhL*MPL*SC
3 0.24 0.65 0.10 0.10 0.24
4 0.00 0.35 0.89 0.89 0.28
5 0.98 0.15 0.33 0.33 0.98
6 0.24 0.00 0.58 0.58 0.24
7 0.00 0.87 0.00 0.00 0.00
8 0.00 0.74 0.46 0.46 0.16
r
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
add a comment |
I have a data frame df1 with four variables. One refers to sunlight, the second one refers to the moon-phase light (light due to the moon's phase), the third one to the moon-position light (light from the moon depending on if it is in the sky or not) and the fourth refers to the clarity of the sky (opposite to cloudiness).
I call them SL, MPhL, MPL and SC respectively. I want to create a new column referred to "global light" that during the day depends only on SL and during the night depends on the other three columns ("MPhL", "MPL" and "SC"). What I want is that at night (when SL == 0), the light in a specific area is equal to the product of the columns "MPhL", "MPL" and "SC". If any of them is 0, then, the light at night would be 0 also.
Since I work with a matrix of hundreds of thousands of rows, what would be the best way to do it? As an example of what I have:
SL<- c(0.82,0.00,0.24,0.00,0.98,0.24,0.00,0.00)
MPhL<- c(0.95,0.85,0.65,0.35,0.15,0.00,0.87,0.74)
MPL<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
SC<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
df<-data.frame(SL,MPhL,MPL,SC)
df
SL MPhL MPL SC
1 0.82 0.95 0.00 0.00
2 0.00 0.85 0.50 0.50
3 0.24 0.65 0.10 0.10
4 0.00 0.35 0.89 0.89
5 0.98 0.15 0.33 0.33
6 0.24 0.00 0.58 0.58
7 0.00 0.87 0.00 0.00
8 0.00 0.74 0.46 0.46
What I would like to get is this:
df
SL MPhL MPL SC GL
1 0.82 0.95 0.00 0.00 0.82 # When "SL">0, GL= SL
2 0.00 0.85 0.50 0.50 0.21 # When "SL" is 0, GL = MPhL*MPL*SC
3 0.24 0.65 0.10 0.10 0.24
4 0.00 0.35 0.89 0.89 0.28
5 0.98 0.15 0.33 0.33 0.98
6 0.24 0.00 0.58 0.58 0.24
7 0.00 0.87 0.00 0.00 0.00
8 0.00 0.74 0.46 0.46 0.16
r
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33
add a comment |
I have a data frame df1 with four variables. One refers to sunlight, the second one refers to the moon-phase light (light due to the moon's phase), the third one to the moon-position light (light from the moon depending on if it is in the sky or not) and the fourth refers to the clarity of the sky (opposite to cloudiness).
I call them SL, MPhL, MPL and SC respectively. I want to create a new column referred to "global light" that during the day depends only on SL and during the night depends on the other three columns ("MPhL", "MPL" and "SC"). What I want is that at night (when SL == 0), the light in a specific area is equal to the product of the columns "MPhL", "MPL" and "SC". If any of them is 0, then, the light at night would be 0 also.
Since I work with a matrix of hundreds of thousands of rows, what would be the best way to do it? As an example of what I have:
SL<- c(0.82,0.00,0.24,0.00,0.98,0.24,0.00,0.00)
MPhL<- c(0.95,0.85,0.65,0.35,0.15,0.00,0.87,0.74)
MPL<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
SC<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
df<-data.frame(SL,MPhL,MPL,SC)
df
SL MPhL MPL SC
1 0.82 0.95 0.00 0.00
2 0.00 0.85 0.50 0.50
3 0.24 0.65 0.10 0.10
4 0.00 0.35 0.89 0.89
5 0.98 0.15 0.33 0.33
6 0.24 0.00 0.58 0.58
7 0.00 0.87 0.00 0.00
8 0.00 0.74 0.46 0.46
What I would like to get is this:
df
SL MPhL MPL SC GL
1 0.82 0.95 0.00 0.00 0.82 # When "SL">0, GL= SL
2 0.00 0.85 0.50 0.50 0.21 # When "SL" is 0, GL = MPhL*MPL*SC
3 0.24 0.65 0.10 0.10 0.24
4 0.00 0.35 0.89 0.89 0.28
5 0.98 0.15 0.33 0.33 0.98
6 0.24 0.00 0.58 0.58 0.24
7 0.00 0.87 0.00 0.00 0.00
8 0.00 0.74 0.46 0.46 0.16
r
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
I have a data frame df1 with four variables. One refers to sunlight, the second one refers to the moon-phase light (light due to the moon's phase), the third one to the moon-position light (light from the moon depending on if it is in the sky or not) and the fourth refers to the clarity of the sky (opposite to cloudiness).
I call them SL, MPhL, MPL and SC respectively. I want to create a new column referred to "global light" that during the day depends only on SL and during the night depends on the other three columns ("MPhL", "MPL" and "SC"). What I want is that at night (when SL == 0), the light in a specific area is equal to the product of the columns "MPhL", "MPL" and "SC". If any of them is 0, then, the light at night would be 0 also.
Since I work with a matrix of hundreds of thousands of rows, what would be the best way to do it? As an example of what I have:
SL<- c(0.82,0.00,0.24,0.00,0.98,0.24,0.00,0.00)
MPhL<- c(0.95,0.85,0.65,0.35,0.15,0.00,0.87,0.74)
MPL<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
SC<- c(0.00,0.50,0.10,0.89,0.33,0.58,0.00,0.46)
df<-data.frame(SL,MPhL,MPL,SC)
df
SL MPhL MPL SC
1 0.82 0.95 0.00 0.00
2 0.00 0.85 0.50 0.50
3 0.24 0.65 0.10 0.10
4 0.00 0.35 0.89 0.89
5 0.98 0.15 0.33 0.33
6 0.24 0.00 0.58 0.58
7 0.00 0.87 0.00 0.00
8 0.00 0.74 0.46 0.46
What I would like to get is this:
df
SL MPhL MPL SC GL
1 0.82 0.95 0.00 0.00 0.82 # When "SL">0, GL= SL
2 0.00 0.85 0.50 0.50 0.21 # When "SL" is 0, GL = MPhL*MPL*SC
3 0.24 0.65 0.10 0.10 0.24
4 0.00 0.35 0.89 0.89 0.28
5 0.98 0.15 0.33 0.33 0.98
6 0.24 0.00 0.58 0.58 0.24
7 0.00 0.87 0.00 0.00 0.00
8 0.00 0.74 0.46 0.46 0.16
r
r
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
edited Mar 28 at 7:33
Cettt
4,9435 gold badges18 silver badges41 bronze badges
4,9435 gold badges18 silver badges41 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
asked Mar 27 at 22:54
DekikeDekike
3029 bronze badges
3029 bronze badges
Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33
add a comment |
Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33
Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33
add a comment |
1 Answer
1
active
oldest
votes
the most simple way would be to use the ifelse function:
GL <- ifelse(SL == 0, MPhL * MPL * SC, SL)
If you want to work in a more structured environment, I can recommend the dplyr package:
library(dplyr)
tibble(SL = SL, MPhL = MPhL, MPL = MPL, SC = SC) %>%
mutate(GL = if_else(SL == 0, MPhL * MPL * SC, SL))
# A tibble: 8 x 5
SL MPhL MPL SC GL
<dbl> <dbl> <dbl> <dbl> <dbl>
1 0.82 0.95 0.00 0.00 0.820000
2 0.00 0.85 0.50 0.50 0.212500
3 0.24 0.65 0.10 0.10 0.240000
4 0.00 0.35 0.89 0.89 0.277235
5 0.98 0.15 0.33 0.33 0.980000
6 0.24 0.00 0.58 0.58 0.240000
7 0.00 0.87 0.00 0.00 0.000000
8 0.00 0.74 0.46 0.46 0.156584
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
add a comment |
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the most simple way would be to use the ifelse function:
GL <- ifelse(SL == 0, MPhL * MPL * SC, SL)
If you want to work in a more structured environment, I can recommend the dplyr package:
library(dplyr)
tibble(SL = SL, MPhL = MPhL, MPL = MPL, SC = SC) %>%
mutate(GL = if_else(SL == 0, MPhL * MPL * SC, SL))
# A tibble: 8 x 5
SL MPhL MPL SC GL
<dbl> <dbl> <dbl> <dbl> <dbl>
1 0.82 0.95 0.00 0.00 0.820000
2 0.00 0.85 0.50 0.50 0.212500
3 0.24 0.65 0.10 0.10 0.240000
4 0.00 0.35 0.89 0.89 0.277235
5 0.98 0.15 0.33 0.33 0.980000
6 0.24 0.00 0.58 0.58 0.240000
7 0.00 0.87 0.00 0.00 0.000000
8 0.00 0.74 0.46 0.46 0.156584
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
add a comment |
the most simple way would be to use the ifelse function:
GL <- ifelse(SL == 0, MPhL * MPL * SC, SL)
If you want to work in a more structured environment, I can recommend the dplyr package:
library(dplyr)
tibble(SL = SL, MPhL = MPhL, MPL = MPL, SC = SC) %>%
mutate(GL = if_else(SL == 0, MPhL * MPL * SC, SL))
# A tibble: 8 x 5
SL MPhL MPL SC GL
<dbl> <dbl> <dbl> <dbl> <dbl>
1 0.82 0.95 0.00 0.00 0.820000
2 0.00 0.85 0.50 0.50 0.212500
3 0.24 0.65 0.10 0.10 0.240000
4 0.00 0.35 0.89 0.89 0.277235
5 0.98 0.15 0.33 0.33 0.980000
6 0.24 0.00 0.58 0.58 0.240000
7 0.00 0.87 0.00 0.00 0.000000
8 0.00 0.74 0.46 0.46 0.156584
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
add a comment |
the most simple way would be to use the ifelse function:
GL <- ifelse(SL == 0, MPhL * MPL * SC, SL)
If you want to work in a more structured environment, I can recommend the dplyr package:
library(dplyr)
tibble(SL = SL, MPhL = MPhL, MPL = MPL, SC = SC) %>%
mutate(GL = if_else(SL == 0, MPhL * MPL * SC, SL))
# A tibble: 8 x 5
SL MPhL MPL SC GL
<dbl> <dbl> <dbl> <dbl> <dbl>
1 0.82 0.95 0.00 0.00 0.820000
2 0.00 0.85 0.50 0.50 0.212500
3 0.24 0.65 0.10 0.10 0.240000
4 0.00 0.35 0.89 0.89 0.277235
5 0.98 0.15 0.33 0.33 0.980000
6 0.24 0.00 0.58 0.58 0.240000
7 0.00 0.87 0.00 0.00 0.000000
8 0.00 0.74 0.46 0.46 0.156584
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
the most simple way would be to use the ifelse function:
GL <- ifelse(SL == 0, MPhL * MPL * SC, SL)
If you want to work in a more structured environment, I can recommend the dplyr package:
library(dplyr)
tibble(SL = SL, MPhL = MPhL, MPL = MPL, SC = SC) %>%
mutate(GL = if_else(SL == 0, MPhL * MPL * SC, SL))
# A tibble: 8 x 5
SL MPhL MPL SC GL
<dbl> <dbl> <dbl> <dbl> <dbl>
1 0.82 0.95 0.00 0.00 0.820000
2 0.00 0.85 0.50 0.50 0.212500
3 0.24 0.65 0.10 0.10 0.240000
4 0.00 0.35 0.89 0.89 0.277235
5 0.98 0.15 0.33 0.33 0.980000
6 0.24 0.00 0.58 0.58 0.240000
7 0.00 0.87 0.00 0.00 0.000000
8 0.00 0.74 0.46 0.46 0.156584
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
edited Mar 27 at 23:08
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
answered Mar 27 at 23:00
CetttCettt
4,9435 gold badges18 silver badges41 bronze badges
4,9435 gold badges18 silver badges41 bronze badges
add a comment |
add a comment |
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Thanks for including sample data and your desired output. What have you tried so far and did you run into any errors (e.g., taking too long, not desired results, etc.)?
– Andrew
Mar 27 at 23:01
I didn't know how to manage this question. I started with this script: df[,5] <- lapply(df$SL, function(x) ifelse(x>0, x, "NA")). My idea was to apply another script after that when SL=0. But then I thought it should be better to ask here how to do it at once with one script.
– Dekike
Mar 28 at 8:06
No worries, this is a good place to ask. For the future, it is good practice to put your attempted solution along with your question. That way, people see that you tried and can address specific reasons why your answer did not work.
– Andrew
Mar 28 at 12:33