size includes NaN values, count does not: In [46]: df = pd.DataFrame({‘a’:[0,0,1,2,2,2], ‘b’:[1,2,3,4,np.NaN,4], ‘c’:np.random.randn(6)}) df Out[46]: a b c 0 0 1 1.067627 1 0 2 0.554691 2 1 3 0.458084 3 2 4 0.426635 4 2 NaN -2.238091 5 2 4 1.256943 In [48]: print(df.groupby([‘a’])[‘b’].count()) print(df.groupby([‘a’])[‘b’].size()) a 0 2 1 1 2 2 Name: … Read more
In C, NAN is declared in <math.h>. In C++, std::numeric_limits<double>::quiet_NaN() is declared in <limits>. But for checking whether a value is NaN, you can’t compare it with another NaN value. Instead use isnan() from <math.h> in C, or std::isnan() from <cmath> in C++.
You can use DataFrame.fillna or Series.fillna which will replace the Python object None, not the string ‘None’. import pandas as pd import numpy as np For dataframe: df = df.fillna(value=np.nan) For column or series: df.mycol.fillna(value=np.nan, inplace=True)
First, let me point out that NaN is a very special value: By definition, it’s not equal to itself. That comes from the IEEE-754 standard that JavaScript numbers draw on. The “not a number” value is never equal to itself, even when the bits are an exact match. (Which they aren’t necessarily in IEEE-754, it … Read more
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(x)) 1000 loops, best of 3: 244 us per loop In [14]: %timeit np.isnan(np.sum(x)) 10000 loops, best of 3: 97.3 us per loop Unlike min, sum doesn’t require branching, which on modern … Read more
The accepted answer is 100% without question WRONG. Not halfway wrong or even slightly wrong. I fear this issue is going to confuse and mislead programmers for a long time to come when this question pops up in searches. NaN is designed to propagate through all calculations, infecting them like a virus, so if somewhere … Read more
just use replace: In [106]: df.replace(‘N/A’,np.NaN) Out[106]: x y 0 10 12 1 50 11 2 18 NaN 3 32 13 4 47 15 5 20 NaN What you’re trying is called chain indexing: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy You can use loc to ensure you operate on the original dF: In [108]: df.loc[df[‘y’] == ‘N/A’,’y’] = np.nan df … Read more
NaN means “Not a Number”. Java Language Specification (JLS) Third Edition says: An operation that overflows produces a signed infinity, an operation that underflows produces a denormalized value or a signed zero, and an operation that has no mathematically definite result produces NaN. All numeric operations with NaN as an operand produce NaN as a … Read more
Assuming your DataFrame is in df: df.Temp_Rating.fillna(df.Farheit, inplace=True) del df[‘Farheit’] df.columns=”File heat Observations”.split() First replace any NaN values with the corresponding value of df.Farheit. Delete the ‘Farheit’ column. Then rename the columns. Here’s the resulting DataFrame:
JavaScript interprets an empty string as a 0, which then fails the isNAN test. You can use parseInt on the string first which won’t convert the empty string to 0. The result should then fail isNAN.