rank

Rank of a matrix in R

由 余生颓废 提交于 2021-01-16 08:46:52
问题 I want to test the rank of a matrix, is there someone who can recommend a package/function in R for this? 回答1: You can try the function qr ("qr", because it performs a QR decomposition): #define a matrix for this example M <- matrix(data = rnorm(12), ncol = 3) #run the function qr() qr(M)$rank #Alternative: load the Matrix package... require(Matrix) #...and run the function rankMatrix() rankMatrix(M)[1] 回答2: http://cran.r-project.org/web/packages/Matrix/Matrix.pdf, page 101 http://cran.r

How to get rank position from @curRank

由 久未见 提交于 2020-12-15 05:44:54
问题 want to call up a ranking position is it posible with @curRank? I created a rank table with SELECT `item`, (`totalrate` / `nrrates`) AS `rank`, @curRank := @curRank + 1 AS `ranking` FROM `rtgitems`, (SELECT @curRank := 0) r WHERE item REGEXP 'Total' ORDER BY (`totalrate` / `nrrates`) DESC I get a table item rank ranking Karla 9.5 1 Kelly 9.3 2 Arian 9.1429 3 in the kelly page i want to call up her ranking position SELECT `item`, (`totalrate` / `nrrates`) AS `rank`, @curRank := @curRank + 1 AS

elimination the linear dependent columns of a non-square matrix in python

由 时光怂恿深爱的人放手 提交于 2020-12-13 03:44:52
问题 I have a matrix A = np.array([[1,1,1],[1,2,3],[4,4,4]]) and I want only the linearly independent rows in my new matrix. The answer might be A_new = np.array([1,1,1],[1,2,3]]) or A_new = np.array([1,2,3],[4,4,4]) Since I have a very large matrix so I need to decompose the matrix into smaller linearly independent full rank matrix. Can someone please help? 回答1: There are many ways to do this, and which way is best will depend on your needs. And, as you noted in your statement, there isn't even a

elimination the linear dependent columns of a non-square matrix in python

由 别说谁变了你拦得住时间么 提交于 2020-12-13 03:41:03
问题 I have a matrix A = np.array([[1,1,1],[1,2,3],[4,4,4]]) and I want only the linearly independent rows in my new matrix. The answer might be A_new = np.array([1,1,1],[1,2,3]]) or A_new = np.array([1,2,3],[4,4,4]) Since I have a very large matrix so I need to decompose the matrix into smaller linearly independent full rank matrix. Can someone please help? 回答1: There are many ways to do this, and which way is best will depend on your needs. And, as you noted in your statement, there isn't even a

Using sort and rank in R on multiple columns

由 邮差的信 提交于 2020-06-17 03:38:06
问题 I’m trying to rank my hospital name by lowest rate for each state. When multiple hospitals have the same rate, the tie should be broken by using the hospital name and sorting it alphabetically. So far I’ve managed to rank it by rate within the state sorting it by hospital name, but I can’t figure out how to break the ties and rank it without skipping numbers This is what I’ve got so far by using the following code: outcome_data <- read.csv("outcome-of-care-measures.csv", na.strings="Not

oracle分析函数Rank, Dense_rank, row_number

由 强颜欢笑 提交于 2020-03-11 18:02:37
一、使用 rownum 为记录排名： 在前面一篇《 Oracle 开发专题之：分析函数》，我们认识了分析函数的基本应用，现在我们再来考虑下面几个问题： ① 对所有客户按订单总额进行排名 ② 按区域和客户订单总额进行排名 ③ 找出订单总额排名前 13 位的客户 ④ 找出订单总额最高、最低的客户 ⑤ 找出订单总额排名前 25% 的客户 按照前面第一篇文章的思路，我们只能做到对各个分组的数据进行统计，如果需要排名的话那么只需要简单地加上 rownum 不就行了吗？事实情况是否如此想象般简单，我们来实践一下。 【 1 】测试环境： SQL > desc user_order; Name Null ? Type ----------------------------------------- -------- ---------------------------- REGION_ID NUMBER ( 2 ) CUSTOMER_ID NUMBER ( 2 ) CUSTOMER_SALES NUMBER 【 2 】测试数据： SQL > select * from user_order order by customer_sales; REGION_ID CUSTOMER_ID CUSTOMER_SALES ---------- ----------- -------------- 5 1

Panda rolling window percentile rank

由 浪子不回头ぞ 提交于 2020-02-26 15:30:36
问题 I am trying to calculate the percentile rank of data by column within a rolling window. test=pd.DataFrame(np.random.randn(20,3),pd.date_range('1/1/2000',periods=20),['A','B','C']) test Out[111]: A B C 2000-01-01 -0.566992 -1.494799 0.462330 2000-01-02 -0.550769 -0.699104 0.767778 2000-01-03 -0.270597 0.060836 0.057195 2000-01-04 -0.583784 -0.546418 -0.557850 2000-01-05 0.294073 -2.326211 0.262098 2000-01-06 -1.122543 -0.116279 -0.003088 2000-01-07 0.121387 0.763100 3.503757 2000-01-08 0

Gradient not defined Tensorflow

由 爱⌒轻易说出口 提交于 2020-02-25 06:36:28
问题 I had asked a question and was implementing the solution when I found out that the operation tf.math.count_nonzero does not have gradient defined. So I tried the following round about method: eps = 1e-6 a = tf.ones((4, 4, 2, 2), tf.float32) h = tf.linalg.svd(a, full_matrices=False, compute_uv=False) cond = tf.less(h, eps) h = tf.where(cond, tf.zeros(tf.shape(h)), h) i = tf.reduce_sum(h, axis=-1) j = h[:, :, 0] rank_mat = tf.multiply(2., tf.ones((4, 4))) cond = tf.not_equal(i, j) rank_mat = tf

Gradient not defined Tensorflow

由 萝らか妹 提交于 2020-02-25 06:35:08
问题 I had asked a question and was implementing the solution when I found out that the operation tf.math.count_nonzero does not have gradient defined. So I tried the following round about method: eps = 1e-6 a = tf.ones((4, 4, 2, 2), tf.float32) h = tf.linalg.svd(a, full_matrices=False, compute_uv=False) cond = tf.less(h, eps) h = tf.where(cond, tf.zeros(tf.shape(h)), h) i = tf.reduce_sum(h, axis=-1) j = h[:, :, 0] rank_mat = tf.multiply(2., tf.ones((4, 4))) cond = tf.not_equal(i, j) rank_mat = tf

Gradient not defined Tensorflow

由 ℡╲_俬逩灬. 提交于 2020-02-25 06:33:56
问题 I had asked a question and was implementing the solution when I found out that the operation tf.math.count_nonzero does not have gradient defined. So I tried the following round about method: eps = 1e-6 a = tf.ones((4, 4, 2, 2), tf.float32) h = tf.linalg.svd(a, full_matrices=False, compute_uv=False) cond = tf.less(h, eps) h = tf.where(cond, tf.zeros(tf.shape(h)), h) i = tf.reduce_sum(h, axis=-1) j = h[:, :, 0] rank_mat = tf.multiply(2., tf.ones((4, 4))) cond = tf.not_equal(i, j) rank_mat = tf
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