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For each line of a Data.Frame, get the column name where a value is TRUE


For each line of a Data.Frame, get the column name where a value is TRUE

By : Gerard Britton
Date : November 17 2020, 11:52 AM
will be helpful for those in need I've got a data.frame L1States_df that looks like this: , Another possibility is
code :
m <- as.matrix(mydf)
replace(NA, row(m)[m], colnames(m)[col(m)[m]])
# [1] NA       "IdleOn" "IdleOn" "IdleOn" "IdleOn" "PumpOn"
match(rowSums(mydf) > 1, TRUE)
# [1] NA NA NA NA NA NA


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Python testing true/false condition on data frame column and returning output in new column

Python testing true/false condition on data frame column and returning output in new column


By : S. Perfees
Date : March 29 2020, 07:55 AM
With these it helps If needed convert the 'week' column datetime dtype using to_datetime then you can just compare the day attribute using dt.day and use this as the condition for np.where:
code :
In [47]:
df['week'] = pd.to_datetime(df['week'])
df['factor'] = np.where(df['week'].dt.day < 7, 'y', 'x')
df

Out[47]:
        week  day  check factor
0 2017-01-08    8  False      x
1 2017-01-15   15  False      x
2 2017-01-22   22  False      x
3 2017-01-29   29  False      x
4 2017-02-05    5   True      y
How to display a column based on a condition that meets true for corresponding column in data.frame in R

How to display a column based on a condition that meets true for corresponding column in data.frame in R


By : Murat özkan
Date : March 29 2020, 07:55 AM
around this issue Below is my data.frame,it contains NA in bonus and increment. , Are you looking for this?
code :
sapply(df[,2:4], function(x) df[which(x == max(x, na.rm = TRUE)),'name'])
salary     bonus     increment(%) 
"BK"       "JK"      "VK" 
Provide column and row names for data.frame in 1 line

Provide column and row names for data.frame in 1 line


By : Z.L Harrison
Date : March 29 2020, 07:55 AM
Does that help I am not quite sure what you are after. But you can re-name row and column names as below using dimnames - this can be extended to multidimensional arrays as well.
code :
df <- data.frame(A=c(1:3), B=c(4:6))

dimnames(df)[[1]] <- row_names_vector
dimnames(df)[[2]] <- col_names_vector
   rownames(df) <-  row_names_vector
   colnames(df) <-  col_names_vector
dimnames(df) <- list(row_names_vector, col_names_vector)
row_names_vector <- letters[1:3]
col_names_vector <- letters[1:2]
dimnames(df) <- list(row_names_vector, col_names_vector)
How do I perform true or false statement, with a calculation, on a data-frame column?

How do I perform true or false statement, with a calculation, on a data-frame column?


By : user1145921
Date : March 29 2020, 07:55 AM
To fix this issue You don't need to use sapply here, you can put it all into ifelse which is already vectorized
code :
a.2$Tr<-ifelse(as.numeric(a.2$K)>=0.8, (a.2$O-a.2$C)*s,  0)
Convert cuDF data frame column to 1 or 0 for “true”/“false” values

Convert cuDF data frame column to 1 or 0 for “true”/“false” values


By : user3117602
Date : March 29 2020, 07:55 AM
will be helpful for those in need I am using RAPIDS (0.9 release) docker container. How can I do the following with RAPIDS cuDF? , You can do this in the same way as with pandas.
code :
import cudf

df = cudf.DataFrame({'a':[0,1,2,3,4]})
df['new'] = df['a'] >= 3
df['new'] = df['new'].astype('int') # could use int8, int32, or int64

# could also do (df['a'] >= 3).astype('int')
df

    a   new
0   0   0
1   1   0
2   2   0
3   3   1
4   4   1
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