nan

I've got a dataset with a big number of rows. Some of the values are NaN, like this: In [91]: df Out[91]: 1 3 1 1 1 1 3 1 1 1 2 3 1 1 1 1 1 NaN NaN NaN 1 3 1 1 1 1 1 1 1 1 And I want to count the number of NaN values in each string, it would be like this: In [91]: list = <somecode with df> In [92]: list Out[91]: [0, 0, 0, 3, 0, 0] What is the best and fastest way to do it? You could first find if element is NaN or not by isnull() and then take row-wise sum(axis=1) In [195]: df.isnull().sum(axis=1) Out[195]: 0 0 1 0 2 0 3 3 4 0 5 0 dtype: int64 And, if you want the output as list, you can In

I'm looking for the fastest way to check for the occurrence of NaN ( np.nan ) in a NumPy array X . np.isnan(X) is out of the question, since it builds a boolean array of shape X.shape , which is potentially gigantic. I tried np.nan in X , but that seems not to work because np.nan != np.nan . Is there a fast and memory-efficient way to do this at all? (To those who would ask "how gigantic": I can't tell. This is input validation for library code.) NPE Ray's solution is good. However, on my machine it is about 2.5x faster to use numpy.sum in place of numpy.min : In [13]: %timeit np.isnan(np.min

This question already has an answer here: Exponentiation with negative base 2 answers I executed the following code: tau <- 0.25 h <- 0.6 * n ^ (-1 / 5) * (4.5 * dnorm(qnorm(tau)) ^ 4 * qnorm(tau) / (2 * (qnorm(tau) ^ 2 + 1)) ^ 2) ^ (1/5), and R keeps producing NaN . However, R actually computes (4.5 * dnorm(qnorm(tau)) ^ 4 * qnorm(tau) / (2 * (qnorm(tau) ^ 2 + 1)) ^ 2) to be equal to -0.003655336 . The weird thing is when I did the following k <- -0.003655336 k ^ (1 / 3) NaN was produced again. You are calculating the cube root of a negative number. Although, as pointed out by @mra68, a real

I'm trying to do a sum of numbers inside div's, so, I did: $(document).ready(function() { var numbers, sumNumbers; $(".item").each(function() { numbers = $(this).children().text(); numbers = +numbers; sumNumbers += numbers; }); console.log(sumNumbers); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="item"> <span class="itemNum">0</span> </div> <div class="item"> <span class="itemNum">2</span> </div> <div class="item"> <span class="itemNum">1</span> </div> But, same converting the numbers from text to number with +numbers is returned NaN,

I have a long list of tuples and want to remove any tuple that has a nan in it using Python. What I currently have: x = [('Recording start', 0), (nan, 4), (nan, 7), ..., ('Event marker 1', 150)] Result I'm looking for: x = [('Recording start', 0), ('Event marker 1', 150)] I've tried use np.isnan and variants of that, but have had no success and keep getting an error: ufunc 'isnan' is not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule "safe" Any suggestions would be appreciated!! You could use list comprehension

IEEE 754 specifies the result of 1 / 0 as ∞ (Infinity). However, IEEE 754 then specifies the result of 0 × ∞ as NaN. This feels counter-intuitive : Why is 0 × ∞ not 0? We can think of 1 / 0 = ∞ as the limit of 1 / z as z tends to zero We can think of 0 × ∞ = 0 as the limit of 0 × z as z tends to ∞. Why does the IEEE standard follow intuition 1. but not 2.? It is easier to understand the behavior of IEEE 754 floating point zeros and infinities if you do not think of them as being literally zero or infinite. The floating point zeros not only represent the real number zero. They also represent

I want to compute the mean, min and max of a series of Managers returns, as follows: ManagerRet <-data.frame(diff(Managerprices)/lag(Managerprices,k=-1)) I then replace return = 0 with NaN since data are extracted from a database and not all the dates are populated. ManagerRet = replace(ManagerRet,ManagerRet==0,NaN) I have the following 3 function > min(ManagerRet,na.rm = TRUE) [1] -0.0091716 > max(ManagerRet,na.rm = TRUE) [1] 0.007565 > mean(ManagerRet,na.rm = TRUE)*252 [1] NaN Why the mean function returns a NaN value while min and max performe calculation properly? Below you can find the