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Use `purrr::map` with k-means


Use `purrr::map` with k-means

By : Jeremy Kendrick
Date : November 20 2020, 03:01 PM
With these it helps The matrix, by itself is a vector with dim attributes. So, when we directly apply map on the matrix, it goes through the each of the individual elements. Instead, place it in a list
code :
list(matrix(1:50, 5) ) %>% 
         map( ~kmeans(x = .x, centers = 2, iter.max = 10))
 matrix(1:50, 5) %>% 
      kmeans(., centers = 2, iter.max = 10)
list(matrix(1:50, 5), matrix(51:100, 5)) %>% 
            map( ~kmeans(x = .x, centers = 2, iter.max = 10))


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does k-means clusterer of apache commons math contains a means method?

does k-means clusterer of apache commons math contains a means method?


By : Megan Balicki
Date : March 29 2020, 07:55 AM
it should still fix some issue The output of the clustering algorithm must at least contain the cluster assignments, i.e. which cluster each point belongs to. If you have that, then the k-means clustering cluster centers are simply given by the mean of the points that belong to each cluster.
Rolling means and applying means at beginning of a series of data

Rolling means and applying means at beginning of a series of data


By : user3457342
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Use right aligment with partial=TRUE, i.e. rollapplyr(..., partial=TRUE) or rollapply(..., align = "right", partial=TRUE). Here we use rollapplyr:
code :
rollapplyr(df$a, 4, mean, partial = TRUE)
Purrr-Fection: In Search of An Elegant Solution to Conditional Data Frame Operations Leveraging Purrr

Purrr-Fection: In Search of An Elegant Solution to Conditional Data Frame Operations Leveraging Purrr


By : Mihir Prajapati
Date : March 29 2020, 07:55 AM
This might help you Really, you want to avoid calling geocode any more than necessary because it's slow and if you're using Google, you only have 2500 queries per day. Thus, it's best to make both columns from the same call, which can be done with a list column, making a new version of the data.frame with do, or a self-join.
1. With a list column
code :
library(dplyr)
library(ggmap)
library(tidyr)    # For `unnest`

       # Evaluate each row separately
df %>% rowwise() %>% 
    # Add a list column. If lon or lat are NA,
    mutate(data = ifelse(any(is.na(c(lon, lat))), 
                         # return a data.frame of the geocoded results,
                         list(geocode(paste(Street, City, State, Zip))), 
                         # else return a data.frame of existing columns.
                         list(data_frame(lon = lon, lat = lat)))) %>% 
    # Remove old columns
    select(-lon, -lat) %>% 
    # Unnest newly created ones from list column
    unnest(data)

## # A tibble: 6 × 6
##                 Street       City    State   Zip       lon      lat
##                  <chr>      <chr>    <chr> <dbl>     <dbl>    <dbl>
## 1        226 W 46th St   New York New York 10036 -73.98670 40.75902
## 2              5th Ave   New York New York 10022 -73.97491 40.76167
## 3          75 Broadway   New York New York 10006 -74.01205 40.70814
## 4          350 5th Ave   New York New York 10118 -73.98566 40.74871
## 5  20 Sagamore Hill Rd Oyster Bay New York 11771 -73.50538 40.88259
## 6 45 Rockefeller Plaza   New York New York 10111 -73.97771 40.75915
       # Evaluate each row separately
df %>% rowwise() %>% 
    # Make a new data.frame from the first four columns and the geocode results or existing lon/lat
    do(bind_cols(.[1:4], if(any(is.na(c(.$lon, .$lat)))){
        geocode(paste(.[1:4], collapse = ' '))
    } else {
        .[5:6]
    }))
df %>% filter(is.na(lon) | is.na(lat)) %>% 
    select(1:4) %>% 
    bind_cols(geocode(paste(.$Street, .$City, .$State, .$Zip))) %>% 
    bind_rows(anti_join(df, ., by = c('Street', 'Zip')))
df %>% filter(is.na(lon) | is.na(lat)) %>% 
    select(1:4) %>% 
    mutate(address = paste(Street, City, State, Zip)) %>%    # make an address column
    mutate_geocode(address) %>% 
    select(-address) %>%    # get rid of address column
    bind_rows(anti_join(df, ., by = c('Street', 'Zip')))

##                 Street       City    State   Zip       lon      lat
## 1              5th Ave   New York New York 10022 -73.97491 40.76167
## 2  20 Sagamore Hill Rd Oyster Bay New York 11771 -73.50538 40.88259
## 3 45 Rockefeller Plaza   New York New York 10111 -73.97771 40.75915
## 4          350 5th Ave   New York New York 10118 -73.98566 40.74871
## 5          75 Broadway   New York New York 10006 -74.01205 40.70814
## 6        226 W 46th St   New York New York 10036 -73.98670 40.75902
df[is.na(df$lon) | is.na(df$lat), c('lon', 'lat')] <- geocode(paste(df$Street, df$City, df$State, df$Zip)[is.na(df$lon) | is.na(df$lat)])
is there a way I can recycle elements of the shorter list in purrr:: map2 or purrr::walk2?

is there a way I can recycle elements of the shorter list in purrr:: map2 or purrr::walk2?


By : Gutoh Morais
Date : March 29 2020, 07:55 AM
hop of those help? You can put both lists in a data frame and let that command repeat your vectors:
code :
input <- data.frame(a = 1:3, b = 4:9)
purrr::map2(input$a, input$b, sum)
Scikit Learn K-means Clustering & TfidfVectorizer: How to pass top n terms with highest tf-idf score to k-means

Scikit Learn K-means Clustering & TfidfVectorizer: How to pass top n terms with highest tf-idf score to k-means


By : user3172641
Date : March 29 2020, 07:55 AM
this will help I am clustering the text data based on TFIDF vectorizer. The code works fine. It takes entire TFIDF vectorizer output as input to the K-Means clustering and generate a scatter plots. Instead I would like to send only top n-terms based on TF-IDF scores as input to the k-means clustering. Is there a way to achieve that ? , use max_features in TfidfVectorizer to consider the top n features
code :
vect = TfidfVectorizer(ngram_range=(1,3),stop_words='english', max_features=n)
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