## Count occurrences of a specific word in Google Spreadsheet

I have some cells with text. I need to count the occurrences of a specific word from those cells.

## probability – Occurrences of diseases, expected number for 5 people.

Let’s say for getting a disease, that we have the frequency of occurrence p, p1 = 0.2, p2 = 0.6, p3 = 0.2, where each stands for a different disease. If 5 people are tested, what is the expected number that these do occur?
I think that this is a Poisson distribution, so I understand that the expected number of occurrence is just 0.2 + 0.6 + 0.2, but the “5 people are tested” makes me wonder if there is an extra step? Do I just multiply the expected number by 5?

## repeat – deletes rows with first occurrences in a matrix / array

Thank you for responding to Stack Overflow!

• Make sure you respond to the question. Give details and share your research!

But to avoid

• Ask for help, clarification or respond to other responses.
• Make statements based on opinion; save them with references or personal experience.

For more information, see our tips on writing correct answers.

## mysql – How do I find the first line of the longest occurrences of repeating numbers?

I have a table with the following structure:

```id,center,shelf_id,occupied,remaining 1,East,1_1_1,1,0 2,East,1_1_2,1,0 3,East,1_1_3,1,4 4,East,1_1_4,1,0 5,East,1_1_5,1,0 6,East,1_2_1,1,3 7,East,1_2_2,1,0 8,East,1_2_3,0,4 9,East,1_2_4,0,4 10,East,1_2_5,0,4 11,East,1_3_1,0,4 12,East,1_3_2,0,4 13,East,1_3_3,0,4 14,East,1_3_4,0,4 ...```

I would like to find the start of the longest chain of repetitive numbers from this position to the end of the array.

For the example above, the mysql statement should return the row ID of 8 because from row 8 to the end of the table, the value in the occupied and remaining columns is 0 and 4 .

Any help in writing it as a select mysql statement would be greatly appreciated!

## applescript – How to get the number of occurrences of words in a string?

Given the text defined on a variable:

``````set textual to "the quick brown fox jumps over the lazy dog"
``````

I am trying to count the number of times a word exists, in this case `the`

I tried but returned zero:

``````return do shell script "grep -ow 'the' " & quoted form of textual & " | wc -l"
``````

and:

``````return do shell script "grep -o 'the' " & quoted form of textual & "| wc -l"
``````

when i try:

``````return do shell script "echo " & quoted form of textual & "grep -ow 'the' | wc -l"
``````

I receive a return from `1` instead of `2`

In an AppleScript `do shell` how do i get the number of times a substring occurs in a past string? When I search to see if this is requested, all I get is the character count. My final goal is to create this as a handler so that I can pass the word and the string so that I can get the counting occurrence of certain words.

## group by – MySQL: count occurrences on different columns in the same query

I have his table:

``````+----+------+---------+
| id | user | upgrade |
+----+------+---------+
|  0 |   A  |   NULL  |
+----+------+---------+
|  1 |   B  |   NULL  |
+----+------+---------+
|  2 |   C  |    A    |
+----+------+---------+
|  3 |   D  |   NULL  |
+----+------+---------+
``````

With a single request, I would like to know the total number of occurrences grouped by user_id but also to know the total number of updates per user.

Something like that:

``````+------+------+---------+
| user |  nb  | nb_upgd |
+------+------+---------+
|  A   |   2  |    1    |
+------+------+---------+
|  B   |   1  |    0    |
+------+------+---------+
|  C   |   1  |    0    |
+------+------+---------+
``````

I tried without success:

``````SELECT user, COUNT(*) AS nb,
COUNT(CASE WHEN t.user_id=upgrade THEN 1 END) AS nb_upgd
FROM table t GROUP BY user
``````

## c ++ – Number of occurrences of the subsequence [1,2,3] in a vector (version 2)

I am trying to implement this exercise but I have no idea how to start; Could someone suggest a way for me?

PRE: receive a vector of integers (formed only by the numbers 1, 2 and 3) and its length

POS: returns the number of occurrences of the subsequence (1,2,3) (consecutive elements of the sequence) in this vector.
Sequences such as 1,2,2,3 or 1,1,2,2,2,2,3,1,2,2,3, etc. must be taken into account, where several numbers 1 can appear together, several 2 and various 3.

Example 1
Entrance: (1,2,3,1,2,2,3)
Exit: 2

Example 2
Entrance (1,2,2,2,2,2,3,3,2,2)
Exit: 2
* /

``````int ocurrencias123Repetidos(int* vector, int largo) {
int contUno = 0;
int contDos = 0;
for (int i = 0; i < largo; i++)
{
if (vector(i) == 1 && vector(i) < vector(i + 1) && vector(i + 1) != 3 && contUno == 0) {
contUno++;
}
if (vector(i) == 2 && vector(i) < vector(i + 1) && contUno == 1) {
contUno++;
}
else if (vector(i + 1) == 1) {
contUno = 0;
}
if (contUno == 2) {
contDos++;
contUno = 0;
}
}
return contDos;
``````

}

## SUM the occurrences (COUNT) of table 2

i have 2 databases with the same columns (id, name) and lots of rows.
I want to make a request separately then make a union. This is because of the extremely large size of the 2 databases.

SELECT id, count (distinguished name)

FROM tabel_1

GROUP BY id

Starting with this query for the two db, I would like to add up the count results based on the id.

Example:

df1 = ((a, 5), (b, 10))

df1 = ((a, 7), (b, 1), (c, 15))

sum = ((a, 12), (b, 11), (c, 15))

## javascript – Number of occurrences in a string

The steps you need the function to take:

• separate the chain into pieces
• create an object
• assign the object a property for each letter
• count how many occurrences

This function might look like this (I have made it optional to count large or small letters, by default not counting):

``````function charCount(str, keepCase) {
if (!keepCase) str = str.toLowerCase();
var obj = {};
for (var i = 0; i < str.length; i++) {
if (!obj(str(i))) obj(str(i)) = 0;
obj(str(i))++;
}
return obj;
}
``````

and the tests would be:

``````var a = charCount("hello");
var b = charCount("AaBbC");
var c = charCount("AaBbC", true); // contando com letras grandes!

console.log(a, b); // Object {h: 1, e: 1, l: 2, o: 1} Object {a: 2, b: 2, c: 1}
console.log(c); // Object {A: 1, a: 1, B: 1, b: 1, C: 1}
``````

## circuits – Count the number of occurrences in NC¹?

Let $$c ∈ ℕ₊$$ be consistent and $$p∈ {0,1 } ^ c$$ a fixed width pattern $$c$$.

Suppose the input length is structured as a list of blocks $$b_1… b_n$$, each block having a width $$c$$; is it possible to calculate the number of equal blocks $$p$$ (i.e., $$bigl lvert {i ) in binary with a family of NC¹ circuits with uniform log space? Why or why not?