nan

问题 I am trying to take the rowwise max (and min) of two columns containing dates from datetime import date import pandas as pd import numpy as np df = pd.DataFrame({'date_a' : [date(2015, 1, 1), date(2012, 6, 1), date(2013, 1, 1), date(2016, 6, 1)], 'date_b' : [date(2012, 7, 1), date(2013, 1, 1), date(2014, 3, 1), date(2013, 4, 1)]}) df[['date_a', 'date_b']].max(axis=1) Out[46]: 0 2015-01-01 1 2013-01-01 2 2014-03-01 3 2016-06-01 as expected. However, if the dataframe contains a single NaN value

问题 I'm trying to prepare heatmap for my data but I have no idea why this error appears. My data: > dput(head(tbl_ready)) structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.370330677123077, 0, 0, 0, 0, 0, 0.53318856142826, 0, 0, 0, 0, 0, 0.217669675587482, 0, 0, 0, 0.79337589572453, 0, 1, 0.0132525790616207, 0, 0, 1, 0.498415470211292, 0.216961707575178, 0.0646831352678839, 0, 0, 0, 0.778625047514492, 0.165974546372072, 0.076951015613392, 0.889894091237216, 0, 0, 1, 0.129806153151281, 0

问题 Wanted to play with std::nan and std::nanf to create nan values with some custom payload (non boxing). However, it really does not work as expected: However, with Visual Studio 2015, the function apparently is not implemented right. The exact sample proposed by cppreference.com produces: nan("1") = nan (7ff8000000000000) nan("2") = nan (7ff8000000000000) Which is not what we expect. Is VS implementation wrong? If not, what would be the right arguments to use to produce 7ff8000000000001 and

问题 I have a NumPy 2D array as shown below: data.dat X1 X2 X3 X4 1 1 1 1 2 2 4 2 3 3 9 3 4 4 16 4 5 5 25 5 6 6 36 6 7 nan 49 7 8 nan 64 8 9 nan 81 nan 10 nan nan nan Now how do I output the last element of each column ignoring nan in the array. I tried without success the code: A[~np.isnan(A)][-1] Code used import numpy as np with open('data.dat', "r") as data: while True: line = data.readline() if not line.startswith('#'): break data_header = [i for i in line.strip().split('\t') if i] A = np

问题 What's the equivalent of the Matlab statement X(isnan(X))=0 in R? Note X is of type of matrix.csr in R. (This is from pkg:SparseM.) 回答1: Are you sure you want to use the matrix.csr class? It is from the SparseM package and as far as I can tell, at least from the package documentation, there are no is.na<- or is.na[ methods. The Matrix-package does document is.na-methods: > library(Matrix);M <- Matrix(1:6, nrow=4, ncol=3, + dimnames = list(c("a", "b", "c", "d"), c("A", "B", "C"))) > stopifnot

问题 When storing double data in my data acquisition project, I identify all "missing" data using std::numeric_limits::quiet_NaN() . However, I'd like to store some extra information to know why the data is "missing" (data transmission lost, bad checksum, no measurement done, internal error....) so I need many different "nan" values in the end. And they must all be identified as NaN by any legacy code ( x!=x ). I see in IEEE 754-1985 that NaN fraction could be "anything except all 0 bits (since

问题 I am trying to parse date like this 2012-12-07T16:18:15+05:30 which I am receiving from database in string format. The parse function I am using is: var jstime = new Date("2012-12-07T16:18:15+05:30"); var h = jstime.getHours(); var m = jstime.getMinutes(); var s = jstime.getSeconds(); var f = "am" if(h >= 12) { f = "pm"; h = h - 12; } if(h == 0) { h = 12; } var str; str = jstime.toDateString(); str = str +"," + h.toString() + ":" + m.toString() + ":" + s.toString() + " " + f.toString();

问题 When I using tensorflow, the loss suddenly turn into nan, just like: Epoch: 00001 || cost= 0.675003929 Epoch: 00002 || cost= 0.237375346 Epoch: 00003 || cost= 0.204962473 Epoch: 00004 || cost= 0.191322120 Epoch: 00005 || cost= 0.181427178 Epoch: 00006 || cost= 0.172107664 Epoch: 00007 || cost= 0.171604740 Epoch: 00008 || cost= 0.160334495 Epoch: 00009 || cost= 0.151639721 Epoch: 00010 || cost= 0.149983061 Epoch: 00011 || cost= 0.145890004 Epoch: 00012 || cost= 0.141182279 Epoch: 00013 || cost

问题 I would like to find where None is found in the dataframe. pd.DataFrame([None,np.nan]).isnull() OUT: 0 0 True 1 True isnull() finds both numpy Nan and None values. I only want the None values and not numpy Nan. Is there an easier way to do that without looping through the dataframe? Edit: After reading the comments, I realized that in my dataframe in my work also include strings, so the None were not coerced to numpy Nan. So the answer given by Pisdom works. 回答1: You could use applymap with a

问题 I have some numerical code that was developed on AMD64 Linux (using LLVM 3.2). I have recently ported it to OSX 10.9 with XCode. It runs fine, but it fails a lot of the unit tests: it seems that some calculations which on Linux return NaN (or -NaN) now return, on OSX, -NaN (or NaN). Can I safely assume that positive and negative NaNs are equivalent and adjust my unit tests to accept either as a success, or is this a sign of something more serious going wrong? 回答1: There is no notion of a