Hi ,
I am trying to run a if loop inside a for loop . The data has a date column which is formatted. When i am trying to execute the code its giving me error. "argument is of length zero". I checked for missing or NA in data but there is none as i am working with sample data for observation max 50.
I have formatted the date column and it looks fine.
Can anyone help,
Thanks
It would help everyone if you provide a reprex - FAQ: What's a reproducible example (`reprex`) and how do I do one?
This will allow us actually solve the problem you have.
Hi Team,
Please find below the code
df1$Date_Created <- as.Date(df1$Date_Created,"%m/%d/%Y")
df2$Availability.Date=as.Date(df2$Availability.Date, "%m/%d/%Y")
df1$lessmonth1 <- as.Date(df1$Date_Created, "%m/%d/%Y") %m-% months(6)
df1$greatmonth1 <- as.Date(df1$Date_Created, "%m/%d/%Y") %m+% months(6)
for (i in 1:nrow(df1)){
count <- 0
for (j in 1:nrow(df2)){
if((df1[i, "Skill"] == df2[j,"Skill"])&
(df1[i, "job"] == df2[j, "jb"] | df1[i, "job"] +1 == df2[j, "jb"])&
((df1[i, "conditions"]) == "L")&
(df1[i, "Work_Location"] == df2[j, "Base.country"])&
(df2[j, "Availability.date"] >= df1[i, "lessmonth1"] & df2[j, "Availability.date"] <= df1[i, "greatmonth1"])
)
count <- count + 1
}
print(count)
df1[i, "out"] <- count
}
## Its failing only at the last filter - Availability date
Error in if ((df1[i, ""] == df2[j, "Primary.Skill.Pool."]) & :
argument is of length zero
I can't know for sure because you are not providing a valid reproducible example (with sample data) but I think your problem is that you cant use [i,""] for subsetting, see this example.
iris[1,""]
#> NULL
iris[1,]
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1 5.1 3.5 1.4 0.2 setosa
Created on 2019-03-13 by the reprex package (v0.2.1)
Andres, I agree with what you're saying, but I'm not getting what line could be creating this specific error.
I can't find if ((df1[i, ""] == df2[j, "Primary.Skill.Pool."]) & ... in the code that OP posted.
Hi @srini,
It is a little hard to tell what you're going for without a reprex, but using a for-loop over rows in a data frame is usually not a good use pattern in R. I think the dplyr code below reproduces your intent, is easier to read, and should run substantially faster because of the vectorization of R functions.
df1 %>%
mutate(row_id = row_number()) %>%
group_by(row_id) %>%
mutate(out = sum(Skill == df2$Skill &
(job == df2$jb | job + 1 == df2$jb) &
conditions == "L" &
Work_Location == df2$Base.country &
Availability.date > lessmonth1 &
greatmonth1 >= Availability.date)) %>%
ungroup()
I'll start by agreeing with @alexkgold: vectorized is the best approach. Go with that.
While the error message doesn't match the provided code, I think @andresrcs is on the right track. The error message argument is of length zero usually means one of two things:
NULL was providedIn this code, both can happen. The loops go through the vectors 1:nrow(df1) and 1:nrow(df2). If either data.frame has no rows, the looped vector would be c(1, 0). And subsetting a data.frame with just 0 for the row and a single column returns a vector of length 0:
iris[0, "Sepal.Length"]
# numeric(0)
You can avoid this by replacing 1:nrow(df1) with seq_len(nrow(df1)). The latter will return an empty vector if df1 has no rows, so the loop's code won't be run.
A NULL can show up if any of the columns named in the code deosn't actually exist in the respective data.frame.
iris[1, "not a column"]
# NULL
Make sure you haven't misspelled a column name and the data is guaranteed to have those names.
Hi Alex,
Thank you for the reply and the code. I understand your point. One issue is that both the datasets doesn't have common primary key. They are totally different. will there be any difference in the code which you have provided.?
Regards,
Srinivas
You can use the merge function to do a cross-product join:
df1 <- data.frame(x = 1:3)
df2 <- data.frame(y = c("a", "b", "c"))
merge(df1, df2, by = NULL)
# x y
# 1 1 a
# 2 2 a
# 3 3 a
# 4 1 b
# 5 2 b
# 6 3 b
# 7 1 c
# 8 2 c
# 9 3 c
Note the argument by = NULL. This prevents merge() from being "helpful" in deciding shared column names are keys for merging.
Hi @alexkgold
Thank you for the reply and the code. I understand your point. One issue is that both the datasets doesn't have common primary key. They are totally different. will there be any difference in the code which you have provided.?
Regards,
Srinivas
Hi @srini,
I don't believe so.
The code counts the number of times the values match, so I don't think primary keys would be relevant.
That said, it's hard to be confident without a full reprex.
Hi @nwerth,
Thank you so much for the clarification. In my situation the datasets (df1) has to look for the filter conditions in df2 and see if the condition is satisfied if yes then it has to do the count (where it counts the number of times the particular condition is satisfied) . will the merge still work?
Regards,
Srinivas
hi @alexkgold
Thank you. Sure I am the dummy data will provide. Meanwhile i tried to run the code you shared it's showing "+" am i missing something here ?
df1 %>%
+ mutate(row_id = row_number()) %>%
+ group_by(row_id) %>%
+ mutate(out = sum(Skill == df2$Skill &
+ ((job == df2$jb | job + 1 == df2$jb) &
+ conditions == "L" &
+Work_Location == df2$Base.country &
+Availability.date > lessmonth1 &
+greatmonth1 >= Availability.date) %>%
+ ungroup()
+
+
Hi @alexkgold
The dummy date is for df1
| skill | conditions | Job cat | work_location | Date created |
|---|---|---|---|---|
| Art | L | 2 | IND | 1/30/2016 |
| science | E | 3 | NZ | 2/27/2017 |
| maths | L | 4 | CHI | 3/20/2018 |
| maths | L | 5 | SWT | 4/22/2017 |
| sciencce | L | 6 | IND | 5/26/2018 |
for df2
| Job cat | Base Country | Availability date | skill |
|---|---|---|---|
| 2 | IND | 1/30/2016 | Art |
| 3 | NZ | 7/22/2017 | science |
| 4 | NZ | 10/30/2018 | maths |
| 5 | SWT | 12/26/2017 | maths |
| 3 | IND | 6/25/2016 | sciencce |
| 2 | IND | 2/21/2016 | maths |
| 3 | IND | 12/21/2015 | maths |
| 3 | IND | 10/21/2015 | sciencce |
| 5 | SWT | 1/22/2017 | maths |
| 5 | SWT | 7/22/2017 | maths |
| 5 | SWT | 11/22/2016 | sciencce |
This is not on a copy/paste friendly format, could you please turn this into a self-contained REPRoducible EXample (reprex)? A reprex makes it much easier for others to understand your issue and figure out how to help.
If you've never heard of a reprex before, you might want to start by reading this FAQ:
Hi @alexkgold,
Hi @andresrcs
Please find below the reproducible code. Please let me know if i missed anything,
library(lubridate)
df1 <- data.frame(Job.cat = c(2L, 3L, 4L, 5L, 6L),
skill = as.factor(c("Art", "science", "maths", "maths", "sciencce")),
conditions = as.factor(c("L", "E", "L", "L", "L")),
work_location = as.factor(c("IND", "NZ", "CHI", "SWT", "IND")),
Date.created = as.factor(c("1/30/2016", "2/27/2017", "3/20/2018",
"4/22/2017", "5/26/2018")))
df2 <- data.frame(Job.cat = c(2L, 3L, 4L, 5L, 3L, 2L, 3L, 3L, 5L, 5L, 5L),
Base.Country = as.factor(c("IND", "NZ", "NZ", "SWT", "IND", "IND", "IND",
"IND", "SWT", "SWT", "SWT")),
Date.Available = as.factor(c("1/30/2016", "7/22/2017", "10/30/2018",
"12/26/2017", "6/25/2016", "2/21/2016",
"12/21/2015", "10/21/2015", "1/22/2017", "7/22/2017",
"11/22/2016")),
skill = as.factor(c("Art", "science", "maths", "maths", "sciencce",
"maths", "maths", "sciencce", "maths", "maths",
"sciencce")))
df1$Date.created <- mdy(df1$Date.created)
df2$Date.Available <- mdy(df2$Date.Available)
# To view the lesser and greater 6 month range
df1$lessmonth <- as.Date(df1$Date.created) %m-% months(6)
df1$greatmonth <- as.Date(df1$Date.created) %m+% months(6)
for (i in 1:nrow(df1)){
count <- 0
for (j in 1:nrow(df2)){
if((df1[i, "skill"] == df2[j,"skill"])&
(df1[i, "job.cat"] == df2[j, "job.cat"] | df1[i, "job.cat"] +1 == df2[j, "job.cat"])&
((df1[i, "conditions"]) == "L")&
(df1[i, "work_location"] == df2[j, "Base.Country"])&
else if ((df1[i, "conditions"]) == "E""] then [(df1[i, "work_location"] == df2[j, "work_location])&
(df2[j, "Date.Available"] >= df1[i, "lessmonth1"] & df2[j, "Date.Available"] <= df1[i, "greatmonth1"])
)
count <- count + 1
}
print(count)
df1[i, "out"] <- count
}
#error Error during wrapup: unexpected 'else' in:
" (df1[i, "work_location"] == df2[j, "Base.Country"])&
else"
Hi Team,
Any help on this will be highly appreciated. Thanks for your time,
Regards,
Srini
I'll suggest you not to expect that others will help you as soon as you post a question. Also, you'll have to remember about the time difference in different parts of the world.
Having said that, you've confused Job.cat with job.cat in your code. Also, you added 1 for both lessmonth and greatmonth inside the loop.
Also, this part is clearly wrong:
There's no if and as far as I know, there's no then in R. This condition was absent in your original code. Are you sure it'll be here? You can't have condition both E and L.
And, please place your codes inside a pair of ```.
library(lubridate)
#>
#> Attaching package: 'lubridate'
#> The following object is masked from 'package:base':
#>
#> date
df1 <- data.frame(Job.cat = c(2L, 3L, 4L, 5L, 6L),
skill = as.factor(c("Art", "science", "maths", "maths", "sciencce")),
conditions = as.factor(c("L", "E", "L", "L", "L")),
work_location = as.factor(c("IND", "NZ", "CHI", "SWT", "IND")),
Date.created = as.factor(c("1/30/2016", "2/27/2017", "3/20/2018", "4/22/2017", "5/26/2018")))
df2 <- data.frame(Job.cat = c(2L, 3L, 4L, 5L, 3L, 2L, 3L, 3L, 5L, 5L, 5L),
Base.Country = as.factor(c("IND", "NZ", "NZ", "SWT", "IND", "IND", "IND", "IND", "SWT", "SWT", "SWT")),
Date.Available = as.factor(c("1/30/2016", "7/22/2017", "10/30/2018", "12/26/2017", "6/25/2016", "2/21/2016", "12/21/2015", "10/21/2015", "1/22/2017", "7/22/2017", "11/22/2016")),
skill = as.factor(c("Art", "science", "maths", "maths", "sciencce", "maths", "maths", "sciencce", "maths", "maths", "sciencce")))
df1$Date.created <- mdy(df1$Date.created)
df2$Date.Available <- mdy(df2$Date.Available)
# To view the lesser and greater 6 month range
df1$lessmonth <- as.Date(df1$Date.created) %m-% months(6)
df1$greatmonth <- as.Date(df1$Date.created) %m+% months(6)
for (i in 1:nrow(df1)){
count <- 0
for (j in 1:nrow(df2)){
if((df1[i, "skill"] == df2[j, "skill"])&
(df1[i, "Job.cat"] == df2[j, "Job.cat"] | df1[i, "Job.cat"] +1 == df2[j, "Job.cat"])&
((df1[i, "conditions"]) == "L")&
(df1[i, "work_location"] == df2[j, "Base.Country"])&
#((df1[i, "conditions"]) == "E") & (df1[i, "work_location"] == df2[j, "work_location"])&
(df2[j, "Date.Available"] >= df1[i, "lessmonth"] & df2[j, "Date.Available"] <= df1[i, "greatmonth"]))
count <- count + 1
}
print(count)
df1[i, "out"] <- count
}
#> [1] 1
#> [1] 0
#> [1] 0
#> [1] 2
#> [1] 0
Hi @Yarnabrina,
Thank you so much for the working code. I completely understand that all are working in different time zone and it's not possible to reply asap.
Yes i understand your point that if and then can't be in used . But i have a filter in which if conditions is L then the work location is same as base country. (For e.g if conditions is "L" and work location is India then only the country with India under Base country is valid. but if the conditions is "E" and the work location is India then any countries under base country can be valid. That's the reason i was using both.
is there any alternative to this?
Thanks and Regards,
Srinivas