Count specific condition in 2d-array

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Given a 2D Array, count the number data at column 3 where it is the same.

For example,

 String[][] array = {
                    { "red", "value1", "alpha"  },
                    { "blue", "value2", "alpha"  }, 
                    { "green", "value3", "alpha"  }, 
                    { "black", "value4", "alpha" },
                    { "grey", "value1", "beta"  },
                    { "white", "value2", "beta"  }, 

          int counter=0;
          for(String[] subArray : array)
              for(String string : subArray)


The output of the counter for beta = 2 and alpha = 4.

Assuming I don't know what is the exact name of the data in column 3, how I count the conditions? if(string.toLowerCase().equals(???))

Any help will be appreciated!

It could be something like this using groupingBy in java 8:
       .collect(Collectors.groupingBy(arr -> arr[2], Collectors.counting()));

Result will:

{alpha=4, beta=2}

B.T.W If you want to ignore case, do it as:
      .collect(Collectors.groupingBy(arr -> arr[2].toLowerCase(), Collectors.counting()));

NumPy: Count the number of elements satisfying the condition, If you want to replace an element that satisfies the conditions, see the following post. In the case of a two-dimensional array, axis=0 gives the count per processed for each row or column when parameter axis is specified. Count dates in a specific date range To count the dates that fall in a certain date range, you can also use either a COUNTIFS formula with two criteria or a combination of two COUNTIF functions. For example, the following formulas count the number of dates in cells C2 through C10 that fall between 1-Jun-2014 and 7-Jun-2014, inclusive:

Create a Map<String, int>

Look for the value of string.toLowerCase in the Map and if it is found increment the value of int else insert with the value of 1

NumPy: Extract or delete elements, rows and columns that satisfy , Extract elements that satisfy the conditions Extract rows and columns that satisfy the post: NumPy: Count the number of elements satisfying the condition Even if the original ndarray is a multidimensional array, a flattened  python .count for multidimensional arrays (list of lists) Simple way to count number of specific elements in 2D array Python Assigning a variable the count of

There are many ways, but one way:

  1. Create a new multidimensional array of types string:int.

  2. Iteratethrough your array at the last index, and then check if it exists in your array you made.

  3. If it exists, inremement the int to int+1.
  4. If it does not exist, add it and add int=1

Sorting, searching, and counting, sort (a[, axis, kind, order]), Return a sorted copy of an array. nanargmax (a[, axis]), Return the indices of the maximum values in the specified axis ignoring NaNs. where (condition, [x, y]), Return elements, either from x or y, depending on  This will work whenever you want to count occurences of a value in arrays of this format. A general and simple answer would be: Since your ndarray contains only 0 and 1, you can use sum() to get the occurrence of 1s and len()-sum() to get the occurrence of 0s. You have a special array with only 1 and 0 here.

numpy.nonzero, Returns a tuple of arrays, one for each dimension of a, containing the indices of the count_nonzero: Counts the number of non-zero elements in the input array. Given an array a, the condition a > 3 is a boolean array and since False is  To count with multiple criteria and OR logic, you can use the COUNTIFS function with an array constant . In the example shown, the formula in H6 is: = SUM ( COUNTIFS ( D4:D11 , { "complete" , "pending" } )) How this formula works

Count rows in a matrix that consist of same element, Given a matrix mat[][], the task is to count the number of rows in the matrix that Approach: Set count = 0 and start traversing the matrix row by row and for a  p31 = numpy.asarray(o31) za = (p31 < 200).sum() # p31<200 is a boolean array, so `sum` counts the number of True elements. Share a link to this answer. improve this answer. answered Oct 21 '12 at 8:32. 222 silver badges. 312 bronze badges. There are many ways to achieve this, like flatten-and-filter or simply enumerate, but I think using

Count all sorted rows in a matrix, Given a matrix of m*n size, the task is to count all the rows in a matrix that are sorted either in strictly increasing order Note c > 1 condition is required to make​. In the C# language, the Length or Count property is often evaluated each time a for-loop is run. It can be faster to hoist the Count or Length check out of the loop. Note: Occasionally the JIT compiler can optimize loops that include the Count or Length in the for statement. Note 2: Hoisting is sometimes faster. For optimal performance, you will need to test your program.

  • Create a new multidimensional array of types string:int How would you know how big to make this array? Easier to use a Map I think
  • True, though, by creating the function yourself you can learn the logic yourself in addition to perhaps expanding using more simpler means. But agreed there are lots of ways.