quantile

问题 I don't understand the following behavior with quantile . With type=2 it should average at discontinuities, but this doesn't seem to happen always. If I create a list of 100 numbers and look at the percentiles, then shouldn't I take the average at every percentile? This behavior happens for some, but not for all (i.e. 7th percentile). quantile(seq(1, 100, 1), 0.05, type=2) # 5% # 5.5 quantile(seq(1, 100, 1), 0.06, type=2) # 6% # 6.5 quantile(seq(1, 100, 1), 0.07, type=2) # 7% # 8 quantile(seq

问题 I have two columns of data representing the same quantity; one column is from my training data, the other is from my validation data. I know how to calculate the percentile rankings of the training data efficiently using: pandas.DataFrame(training_data).rank(pct = True).values My question is, how can I efficiently get a similar set of percentile rankings of the validation data column relative to the training data column? That is, for each value in the validation data column, how can I find

问题 I used the quantreg package in R to compute the quantile regression model. In the model, dependent Variable(Y) is NAS_DELAY , and the independent variable(Xs) are SEANSON1TO4 , SEANSON2TO4 , SEANSON3TO4 . The model is: NAS_DELAY=aSEANSON1TO4+bSEANSON2TO4+cSEANSON3TO4+d The SEANSON1TO4 , SEANSON2TO4 , SEANSON3TO4 are dummy variables, 0 or 1. I use R to compute the intercept and other regression coefficient, but the result showed that "error in rq.fit.br(x,y,tau=tau,....)singular design matrix

问题 I'm giving a sorted array of 24 numbers to d3.quantile and asking it to calculate the first quartile value. Since the array can be split evenly into four groups of 6 values, my assumption was that the result would be the mean of arr[5] and arr[6], but that's not what I got. var arr = [89.7, 93.2, 94, 94.3, 94.5, 95.4, 95.9, 96.1, 96.4, 96.5, 96.9, 96.9, 97.3, 97.6, 97.6, 97.6, 97.8, 98.3, 98.3, 98.4, 98.5, 98.5, 98.6, 98.6]; var myAssumption = (arr[5] + arr[6]) / 2; // 95.65 var d3Result = d3

问题 I’m analyzing river streamflow data with R language and I have two nested lists. First holds data (Flowtest) from different river reaches called numbers such as 910, 950, 1012 and 1087. I have hundreds of daily streamflow measurements (Flow), but as I’m preparing yearly statistics the exact day and month doesn’t matter. Each measurement (Flow) is referenced to a year (Year) in the Flowtest table. Flowtest <- list("910" = tibble(Year = c(2004, 2004, 2005, 2005, 2007, 2008, 2008), Flow=c(123,

问题 I’m analyzing river streamflow data with R language and I have two nested lists. First holds data (Flowtest) from different river reaches called numbers such as 910, 950, 1012 and 1087. I have hundreds of daily streamflow measurements (Flow), but as I’m preparing yearly statistics the exact day and month doesn’t matter. Each measurement (Flow) is referenced to a year (Year) in the Flowtest table. Flowtest <- list("910" = tibble(Year = c(2004, 2004, 2005, 2005, 2007, 2008, 2008), Flow=c(123,

问题 Consider I have a million of observations following Gamma distribution with parameters (3,5). I am able to find the quantiles using summary() but I am trying to find how many observations are between each red lines which were divided into 10 pieces? a = rgamma(1e6, shape = 3, rate = 5) summary(a) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0053 0.3455 0.5351 0.6002 0.7845 4.4458 回答1: We may use cut with table : table(cut(a, quantile(a, 0:10 / 10))) # (0.00202,0.22] (0.22,0.307] (0.307,0.382] (0

问题 Consider I have a million of observations following Gamma distribution with parameters (3,5). I am able to find the quantiles using summary() but I am trying to find how many observations are between each red lines which were divided into 10 pieces? a = rgamma(1e6, shape = 3, rate = 5) summary(a) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0053 0.3455 0.5351 0.6002 0.7845 4.4458 回答1: We may use cut with table : table(cut(a, quantile(a, 0:10 / 10))) # (0.00202,0.22] (0.22,0.307] (0.307,0.382] (0

问题 The quantile functions gives us the quantile of a given pandas series s , E.g. s.quantile(0.9) is 4.2 Is there the inverse function (i.e. cumulative distribution) which finds the value x such that s.quantile(x)=4 Thanks 回答1: I had the same question as you did! I found an easy way of getting the inverse of quantile using scipy. #libs required from scipy import stats import pandas as pd import numpy as np #generate ramdom data with same seed (to be reproducible) np.random.seed(seed=1) df = pd

问题 The quantile functions gives us the quantile of a given pandas series s , E.g. s.quantile(0.9) is 4.2 Is there the inverse function (i.e. cumulative distribution) which finds the value x such that s.quantile(x)=4 Thanks 回答1: I had the same question as you did! I found an easy way of getting the inverse of quantile using scipy. #libs required from scipy import stats import pandas as pd import numpy as np #generate ramdom data with same seed (to be reproducible) np.random.seed(seed=1) df = pd