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Questions - min

Create a table from dataframe column values mean and standard deviation in R

I am new to R so no idea about the code. I have two data frames. One dataframe looks like this. df ID Disease GSM239170 Control GSM239323 Control GSM239324 Control GSM239326 Control GS...
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user17416440

r

function

max

mean

min

Votes: 0

Answers: 4

Latest Answer

We could combine pivot_longer with right_join and then use summarise on the group: library(dplyr) library(tidyr) df1 %>% pivot_longer( everything(), names_to = "ID", value...
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TarJae

Incremental algorithm for maximum of last N entries in a series

Does anyone know of an efficient incremental algorithm for the maximum (or minimum) of the last n entries of an array? By "efficient" and "incremental" I mean that when a new eleme...
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barneypitt

algorithm

max

min

Votes: 0

Answers: 3

Latest Answer

You are lucky, there is a cool algorithm due to van Herk, Gil and Werman that computes a 1D dilation in constant time per element. https://tpgit.github.io/UnOfficialLeptDocs/leptonica/grayscale-morpho...
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user1196549

Ignore NA values when using MIN function of "apply"

I am trying to get a column made up of the minimum values from other specified columns in which NA values are ignored. When I do this: foo = data.frame(A = c(1, 2, 3, NA), B = c(3, NA, 2, NA), C = c(N...
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Sergei Walankov

r

dataframe

apply

min

na

Votes: 0

Answers: 1

Latest Answer

You can add na.rm=T to your function and set to NAwhen the entire vector of values is NA (otherwise it returns it returns -Inf). my_min <- function(x) ifelse( !all(is.na(x)), min(x, na.rm=T), NA) f...
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Ma&#235;l

How can I find all datetimes in a list within n minutes of a given value?

I have two lists of times (Hour:Min:Sec format) and I've been struggling to compare each entry in list_a against all of list_b to identify values that fall within 30 minutes: list_a = ["10:26:42&...
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Blumie

python

list

datetime

min

python-datetime

Votes: 0

Answers: 4

Latest Answer

You loop and loot at the absolute time diff of all elements instead of using min: list_a = ["10:26:42", "8:55:43", "7:34:11"] list_b = ["10:49:20", "8:51:1...
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Patrick Artner

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