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How to format dplyr output in R into doubles (or other workable format)?


How to format dplyr output in R into doubles (or other workable format)?

By : Gianina
Date : November 22 2020, 03:01 PM
hope this fix your issue How do I get the outcome of this to be numerical values versus tibbles. , coerce it to a dataframe.
code :
prop_smoking <- as.data.frame(prop_smoking)


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Convert a math equation in string format to something workable in Python3

Convert a math equation in string format to something workable in Python3


By : md waseem
Date : March 29 2020, 07:55 AM
around this issue Use the eval function , and use a list comprehension. This requires you know the var name ahead of time. If you don't, parse it then use this.
code :
>>> operation = "x**4"
>>> num_list = [4,3,2]
>>> [eval(operation) for x in num_list]
[256, 81, 16]
R - transform output format of dplyr summarise

R - transform output format of dplyr summarise


By : Maxim
Date : March 29 2020, 07:55 AM
it fixes the issue You could use spread from tidyr to extend your pipeline. Note, I assigned a name for the mean so it could be referenced simply within the spread call.
code :
library(dplyr)
library(tidyr)

mtcars %>% 
    group_by(gear, vs) %>% 
    summarise(mean_disp = mean (disp) ) %>%
    spread(vs, mean_disp)

Source: local data frame [3 x 3]

  gear        0        1
1    3 357.6167 201.0333
2    4 160.0000 115.6200
3    5 229.3250  95.1000
Format dplyr do() output by group into data.frame

Format dplyr do() output by group into data.frame


By : kavwoman
Date : March 29 2020, 07:55 AM
With these it helps I use dplyr to interpolate different length curves to identical length curves with the do() function. , We can use unnest
code :
library(tidyr)
unnest(df, elapsed_perc, duration_prog)
#      id elapsed_perc duration_prog
#     <chr>        <dbl>         <dbl>
#1      a          0.2          0.12
#2      a          0.4          0.28
#3      a          0.6          0.52
#4      a          0.8          0.72
#5      a          1.0          1.00
#6      b          0.2          0.19
#7      b          0.4          0.48
#8      b          0.6          0.76
#9      b          0.8          0.92
#10     b          1.0          1.00
Reformat strings into a workable data format

Reformat strings into a workable data format


By : user2422554
Date : March 29 2020, 07:55 AM
this one helps. You could use map and reduce to first create array of objects and then one more reduce to get top salary object for each name.
code :
const list = [
  "Record - Name: Peter - Salary: 100000 - Position: Accountant - Date: February 15, 2018, 1:15 PM",
  "Record - Name: Peter - Salary: 120000 - Position: Accountant - Date: February 15, 2019, 1:15 PM",
  "Record - Name: Jonny - Salary: 90000 - Position: Developer - Date: February 15, 2019, 1:15 PM",
  "Record - Name: Peter - Salary: 100000 - Position: Accountant - Date: February 15, 2018, 1:15 PM"
];

const data = list.map(rec => {
  return rec.replace(/ /g, '').split('-').slice(1)
    .reduce((r, e) => {
      let [key, value] = e.split(':')
      r[key.toLowerCase()] = value
      return r;
    }, {})
})

const filtered = data.reduce((r, e) => {
  if (!r[e.name]) r[e.name] = e;
  else if (r[e.name].salary < e.salary) r[e.name] = e
  return r
}, {})

console.log(Object.values(filtered))
Converting separate columns for mm, dd, and yr into a workable mm/dd/year format

Converting separate columns for mm, dd, and yr into a workable mm/dd/year format


By : Nidhish Singhal
Date : March 29 2020, 07:55 AM
it helps some times Basically I have 3 separate columns in a table. I will call them SMonth, Sday, Syear. They are stored as numeric values for some reason. I can use the following string to format them into what looks like a date but doesn't allow me to use functions such as sort, order by, datediff or dateadd. , Just convert your result into a date or datetime.
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