Count the Number of Non-Zero Elements of Each Column

Count the number of non-zero elements of each column

What about:

apply(your.matrix, 2, function(c)sum(c!=0))

Does this help?

edit:

Even better:

colSums(your.matrix != 0)

edit 2:

Here we go, with an example for ya:

> example = matrix(sample(c(0,0,0,100),size=70,replace=T),ncol=7)
> example
      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
 [1,]    0  100    0    0  100    0  100
 [2,]  100    0    0    0    0    0  100
 [3,]    0    0    0    0    0    0  100
 [4,]    0  100    0    0    0    0    0
 [5,]    0    0  100  100    0    0    0
 [6,]    0    0    0  100    0    0    0
 [7,]    0  100  100    0    0    0    0
 [8,]  100    0    0    0    0    0    0
 [9,]  100  100    0    0  100    0    0
[10,]    0    0    0    0    0  100    0
> colSums(example != 0)
[1] 3 4 2 2 2 1 3

(new example, the previous example with '1' values was not suited to show that we are summing the number of cells, not their contents)

Count the number of non-zero elements of each column- Piping Friendly

example %>% summarise(across(where(is.numeric), ~sum(. != 0)))

count nonzero values in each column tidyverse

A possible solution:

library(dplyr)

df %>% 
  summarise(across(starts_with("site"), ~ sum(.x != 0)))

#> # A tibble: 1 × 3
#>   site1 site2 site3
#>   <int> <int> <int>
#> 1     2     3     1

---

Another possible solution, in base R:

apply(df[-1], 2, \(x) sum(x != 0))

#> site1 site2 site3 
#>     2     3     1

calculating count of non zero values in row in dataframe

You can use the regex 'T_1_[1-9]' to exclude T_1_0.

library(dplyr)
df <- df %>% mutate(new1 = rowSums(select(., matches('T_1_[1-9]')) !=0))
df
#  T_1_1 T_1_2 T_1_3 S_2_1 S_2_2 S_2_3 T_1_0 new1
#1    68    26    93    69    87   150    79    3
#2    24   105    32    67    67     0     0    3
#3     0     0    73    94    47   132     0    1
#4   105    73   103     0   120   121   115    3
#5    58    39   149    77    85    29    98    3
#6     0    97     0   136   122    78    12    1
#7   135    46   147   137     0   109    15    3
#8   126   108   113    92    96    40   121    3
#9    24     0   139    73    79     0     2    2

Or more specific :

df <- df %>% mutate(new1 = rowSums(select(., starts_with('T_1'), -T_1_0) !=0))

Count the number of column having non zero values

SELECT COUNT(*) AS Total,
       ...
       ... 
       CASE ( WHEN SUM(CASE WHEN TypeId ='4' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END) +  
       CASE ( WHEN SUM(CASE WHEN TypeId ='6' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END) + 
       CASE ( WHEN SUM(CASE WHEN TypeId ='1' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END) + 
       CASE ( WHEN SUM(CASE WHEN TypeId ='10' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END) + 
       CASE ( WHEN SUM(CASE WHEN TypeId ='5' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END) + 
       CASE ( WHEN SUM(CASE WHEN TypeId ='8' THEN 1 ELSE 0 END) > 0 THEN 1 ELSE 0 END)  
          as SumOfNonZeros

FROM [Party]

Or maybe simpler

SELECT COUNT(*) AS Total,
       COUNT(CASE WHEN TypeId ='4'  THEN 1 END) AS 'TotalCount1', 
       COUNT(CASE WHEN TypeId ='6'  THEN 1 END) AS 'TotalCount2', 
       COUNT(CASE WHEN TypeId ='1'  THEN 1 END) AS 'TotalCount3', 
       COUNT(CASE WHEN TypeId ='10' THEN 1 END) AS 'TotalCount4', 
       COUNT(CASE WHEN TypeId ='5'  THEN 1 END) AS 'TotalCount5', 
       COUNT(CASE WHEN TypeId ='8'  THEN 1 END) AS 'TotalCount6',
       COUNT( DISTINCT CASE WHEN TypeId IN ('4', '6', '1', '10', '5', '8') 
                            THEN TypeId 
                       END ) as CountOfNonZeros
FROM [Party]

Count non-zero entries in each column of a matrix

It's simply

> A ~= 0
ans =

   1   1   1   1   1
   1   1   1   1   1
   0   0   1   0   1

> sum(A ~= 0, 1)
ans =

   2   2   3   2   3

Count non zero entry in row in R

In base R

df$count <- rowSums(df!=0)
  mode1 mode2 mode3 count
1     8     1     0     2
2     0     0     0     0
3     6     5     4     3
4     1     2     3     3
5     1     1     1     3

Using dplyr

library(dplyr)
df %>% mutate(count=rowSums(.!=0))
#For specific columns we can use
df %>% mutate(count=rowSums(.[1:3]!=0))
#OR
df %>% mutate(count=rowSums(select(.,starts_with("mode"))!=0))

Count the number of non-zero columns in a matrix

If you want the columns that have non-zero elements, meaning they may have zeros but not all the rows are zeros then the following works:

mat<-matrix(rep(0,2*5),ncol=2)
mat[,1]=c(1,2,3,4,5)

sum(colSums(mat!=0)!=0)
#> [1] 1

But if you want to find the column which have no zeros, look below. I am making an example to illustrate that better:

mat2<-matrix(rep(0,4*5),ncol=4)
mat2[,1]=c(0,2,3,4,5)
mat2[,2]=c(1,2,3,4,5)
mat2[,3]=c(0,0,0,-1,1)

sum(colSums(mat2!=0)!=0) #count of non-zero columns
#> [1] 3

sum(colSums(mat2!=0)==nrow(mat2)) #count of columns with no zeros
#> [1] 1

If you want to find out which columns are non-zero then use which instead of sum:

which(colSums(mat2!=0)!=0) #non-zero columns
#> [1] 1 2 3

which(colSums(mat2!=0)==nrow(mat2)) #columns with no zeros
#> [1] 2

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