Replace columns less than a threshold to 0

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I have a dataframe that has 200 columns. I want to use dplyr to clean the data so that every number less than 0.05 is replaced with 0. A sample df was pasted below.

df
0.07262
0.039885
0.090173
0.124043
0.09201
0.068309
0.146381
0.09127
0.060768
0.111031

This is the desired outcome.

df
0.07262
0
0.090173
0.124043
0.09201
0.068309
0.146381
0.09127
0.060768
0.111031

This is my code: df2 <- mutate_all(ifelse(<0.05,0.,)) but it doesn't work. Any guidance is welcome.

dplyr:

df <- mutate_all(df, funs(ifelse(. < 0.5, 0, .)))

base R:

df[df < 0.05] <- 0

Replace 'greater than' values in a matrix, 0. Accepted Answer: Andrei Bobrov. I want to look through a very large matrix and replace numbers that are larger than, say, a, with b. Is there a simple way or a  For example, I will find all the values which greater than 500, and replace them with 0. The following VBA code can help you to replace the values which greater or less than a specific value at one, please do as follows: 1. Hold down the ALT + F11 keys, and it opens the Microsoft Visual Basic for Applications window. 2.

Here's one way using base R

  df[df < 0.05] <- 0

How to replace values less than zero in a cell array with NaN , I have a huge cell array and there is a column with a lot of negative values. I want to clean up and replace all those values with NaN. Would appreciate any help. Query editor replace values less than threshold. Subscribe to RSS Feed. Email to a Friend. Report Inappropriate Content. ‎08-10-2017 03:31 AM. Hi, is it possible to replace values in a column that are less than e.g. 1 to null in Query editor? Solved! Go to Solution. Message 1 of 10. 2 ACCEPTED SOLUTIONS. Accepted Solutions.

Using dplyr, you can define the function before and then call it in the mutate_all

smallToZero <- function(x) {if_else(x<.05, 0, x)}
df2 <- 
  df %>% 
  mutate_all(smallToZero)

You can also have it as an un-named function using . as:

df2 <- 
  df %>% 
  mutate_all(funs(if_else(.<.05, 0, .)))

Solved: Query editor replace values less than threshold, Solved: Hi, is it possible to replace values in a column that are less than e.g. 1 to null in Query editor? Re: Query editor replace values less than threshold ReplaceValue(#"Sorted Rows1",each {"Pumptime"},each if {"Pumptime"} < 0 then​  I have a 3000 x 3000 matrix and want to remove all elements inside the matrix that are less than a specific value, .5 for example.

Pattern Recognition in Bioinformatics: 4th IAPR International , (2) ⎪⎭ ⎪⎩ else 0 3.2 Construction of Unweighted Graph In order to map the by replacing all values greater than the entropy threshold to 0 and those less than  Replace 'greater than' values in a matrix Accepted Answer: Andrei Bobrov. I want to look through a very large matrix and replace numbers that are larger than, say

7.3 Changing values of a vector, 8.4.1 Adding new columns · 8.4.2 Changing column names We'll index the values 6 through 10, and assign a value of 0. a[6:10] <- 0 a As you can see, our new values of x are now never less than 1 or greater than 10! In other languages this code wouldn't work because we're trying to replace 5 values with just 1. Write a NumPy program to replace all elements of NumPy array that are greater than specified array. Resetting will undo all of your current changes. pro tip You can save a copy for yourself with the Copy or Remix button. Recaptcha requires verification. pro tip You can save a copy for yourself with the Copy or Remix button. Publish Your Trinket!

Conditionally Altering Array Elements, The following statement sets elements of A with values of zero or less to -1: The following statements replace all drop-outs with the average of the two adjacent  Value to replace any values matching to_replace with. For a DataFrame a dict of values can be used to specify which value to use for each column (columns not in the dict will not be filled). Regular expressions, strings and lists or dicts of such objects are also allowed. inplace bool, default False. If True, in place.

Comments
  • You need to use dplyr or is base R enough?