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How to remove rows if a list of columns is completely full of NAs (not removing the rows with at least one value other t


By : terron
Date : September 17 2020, 08:00 AM
Any of those help You can use grep to find columns with rev and use all with apply to find rows where all are NA.
code :
dat[!apply(is.na(dat[,grep("^rev", colnames(dat))]), 1, all),]
#  Company survey_year rev_01 rev_02 rev_03 rev_04 rev_05 variable
#1       a        2014     NA     10     20     NA     NA        U
#2       b        2010     20     50     NA     NA     30        P
#4       d        2014     NA     30     NA     50     NA        E
#6       f        2010     10     20     30     50     60        T
dat[rowSums(!is.na(dat[,grep("^rev", colnames(dat))])) > 0,]


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How can I remove rows containing '0' of certain columns while keeping the rows IDs of remaining rows in R


By : eulaliefow
Date : March 29 2020, 07:55 AM
it helps some times if you want to remove the lines containing a 0 or many for column B,C or D :
code :
DF[apply(DF[c(2:4)],1,function(z) !any(z==0)),] 
DF[apply(DF[c(2:4)],1,function(z) any(z!=0)),]

How to remove duplicate ID rows. While removing, use the rows that has NULL value in another column


By : Dynamics
Date : March 29 2020, 07:55 AM
This might help you Assuming you have SQL Server 2008 or above, this will work for you. I use row_number and assign the values by ID starting at the max date. So any value higher than 1 is lower than the max date for that particular ID so I delete row_num greater than 1.
Check it out:
code :
DECLARE @yourTable TABLE (ID INT,Sale_date DATE, Price FLOAT);

INSERT INTO @yourTable
VALUES  (11,'20051020',22.1),
        (11,NULL,20.1),
        (12,NULL,20.1),
        (13,'20051020',20.1);

WITH CTE
AS
(
    SELECT  *,
            ROW_NUMBER() OVER (PARTITION BY ID ORDER BY sale_date DESC) AS row_num
    FROM @yourTable
)

DELETE
FROM CTE
WHERE row_num > 1

SELECT *
FROM @yourTable

Removing specific columns and rows from nested list


By : passeddust
Date : March 29 2020, 07:55 AM
This might help you If for example I have glider = [[0,0,0,0],[1,2,3,4],[0,1,3,4],[0,0,0,0]], how would I go about deleting the first and last list as well as the first and last character per list if the nested lists varies. It would look like this after. , You can do this using slicing and a temporary list:
code :
glider = [[0,0,0,0],[1,2,3,4],[0,1,3,4],[0,0,0,0]]

glider = glider[1:-1]

templist = []

for i in glider:
    templist.append(i[1:-1])
glider = templist
del templist
>>> glider
[[2, 3], [1, 3]]

R remove rows with most zero values (unique and removing all rows with 0 not working)


By : user2695015
Date : March 29 2020, 07:55 AM
Does that help Hi I am stuck on a weird problem. , You could use dplyr for that as below:
code :
library(dplyr)
Final_data$CountZero <- apply(Final_data[, -(1:2)], 1, function(x) {
  sum(x == 0)
})

Final_data %>%
  group_by(City, Grade) %>%
  filter(CountZero == min(CountZero)) %>%
  select(-CountZero)
# A tibble: 2 x 5
# Groups:   City, Grade [2]
  City  Grade Variable1 Variable2 Variable3
  <fct> <fct>     <dbl>     <dbl>     <dbl>
1 XX    A          0.34      0.76      0.76
2 YY    B          0         0.3       0.3 

Removing all empty columns and rows in data.frame when rows don't go away


By : user3265786
Date : March 29 2020, 07:55 AM
This might help you I have two data.frames, dA (HERE) and dB (HERE). They are exactly the same EXCEPT that dB has one completely empty column and multiple empty rows. , You have NA and also empty rows. You can do
code :
B1[rowSums(is.na(B1) | B1 == "") != ncol(B1), ]

#   study.name  group.name outcome ESL prof scope type
#1  Shin.Ellis    ME.short       1   1    2     1    1
#2  Shin.Ellis     ME.long       1   1    2     1    1
#3  Shin.Ellis   DCF.short       1   1    2     1    2
#4  Shin.Ellis    DCF.long       1   1    2     1    2
#5  Shin.Ellis  Cont.short       1   1    2    NA   NA
#6  Shin.Ellis   Cont.long       1   1    2    NA   NA
#8    Trus.Hsu       Exper       1   2    2     2    1
#.....
library(dplyr)
B1 %>% filter_all(any_vars(!is.na(.) & . != ""))
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