quartile
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quartile [2019/09/16 11:52] – hkimscil | quartile [2023/09/11 08:42] (current) – [r method] hkimscil | ||
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사분범위 = (상한사분위수) - (하한사분위수) | 사분범위 = (상한사분위수) - (하한사분위수) | ||
- | ---- | ||
====== Finding lower and upper quartile ====== | ====== Finding lower and upper quartile ====== | ||
- | ===== head first ===== | + | ===== e.g. 1, Head First method |
+ | < | ||
+ | > k | ||
+ | [1] 1 2 3 4 5 6 7 8 | ||
+ | > quantile(k) | ||
+ | 0% 25% 50% 75% 100% | ||
+ | 1.00 2.75 4.50 6.25 8.00 | ||
+ | > </ | ||
+ | |||
+ | < | ||
+ | head first | ||
* 하한 | * 하한 | ||
* n / 4 = ? | * n / 4 = ? | ||
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* 정수가 아니면? 올림을 한 위치 값 | * 정수가 아니면? 올림을 한 위치 값 | ||
- | < | + | 위의 방법으로는 |
- | > k | + | |
- | [1] 1 2 3 4 5 6 7 8 | + | |
- | > quantile(k) | + | |
- | 0% 25% 50% 75% 100% | + | |
- | 1.00 2.75 4.50 6.25 8.00 | + | |
- | > </ | + | |
- | 그러나, | + | |
lower quartile: 2.5 | lower quartile: 2.5 | ||
upper quartile: 6.5 | upper quartile: 6.5 | ||
Line 41: | Line 43: | ||
* upper: 43 | * upper: 43 | ||
+ | |||
+ | ===== r method ===== | ||
in r | in r | ||
- | < | + | < |
+ | j <- c(1,2,3,4,5) | ||
+ | j <- sort(j) | ||
+ | quantile(j) | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | > j <- c(1,2,3,4,5) | ||
+ | > j <- sort(j) | ||
> quantile(j) | > quantile(j) | ||
0% 25% 50% 75% 100% | 0% 25% 50% 75% 100% | ||
- | 6.0 25.5 40.0 42.5 49.0 </ | + | |
+ | > | ||
+ | </ | ||
+ | Odd number of elements | ||
+ | * Use the median to divide the ordered data set into two halves. | ||
+ | * If there are an odd number of data points in the original ordered data set, include the median (the central value in the ordered list) in both halves. (가운데 숫자) | ||
+ | |||
+ | |||
+ | < | ||
+ | j2 <- c(1, | ||
+ | j2 <- sort(j2) | ||
+ | quantile(j2) | ||
+ | </ | ||
+ | < | ||
+ | > j2 <- c(1, | ||
+ | > j2 <- sort(j2) | ||
+ | > quantile(j2) | ||
+ | | ||
+ | 1.00 2.25 3.50 4.75 6.00 | ||
+ | > | ||
+ | > | ||
+ | </ | ||
+ | |||
+ | Even number of elements | ||
+ | * If there are an even number of data points in the original ordered data set, split this data set exactly in half. 즉, 3과 4의 가운데 값 (50%) = 3.5 | ||
+ | * lower bound (lower quartile) 앞부분을 반으로 쪼갯을 때의 숫자 (여기서는 2) 더하기, 그 다음숫자와의 차이의 (3-2) 1/4지점 (여기서는 2 + 0.25 = 2.25) 구한다. | ||
+ | * upper bound는 뒷부분의 반인 | ||
+ | |||
+ | < | ||
+ | > j3 <- c(7, 18, 5, 9, 12, 15) | ||
+ | > j3s <- sort(j3) | ||
+ | > j3s | ||
+ | [1] 5 7 9 12 15 18 | ||
+ | > quantile(j3s) | ||
+ | 0% | ||
+ | | ||
+ | > | ||
+ | </ | ||
+ | median = (9+12)/2 | ||
+ | the 1st quartile = 7 + (9-7)*(1/4) = 7 + 0.5 = 7.5 | ||
+ | the 3rd quartile = 12 + (12-9)*(3/ | ||
---- | ---- |
quartile.1568602345.txt.gz · Last modified: 2019/09/16 11:52 by hkimscil