I was a member of the IEEE-754 committee, I’ll try to help clarify things a bit. First off, floating-point numbers are not real numbers, and floating-point arithmetic does not satisfy the axioms of real arithmetic. Trichotomy is not the only property of real arithmetic that does not hold for floats, nor even the most important. … Read more
var s=””; var num = parseInt(s) || 0; When not used with boolean values, the logical OR || operator returns the first expression parseInt(s) if it can be evaluated to true, otherwise it returns the second expression 0. The return value of parseInt(”) is NaN. NaN evaluates to false, so num ends up being set … Read more
df = df.fillna(”) This will fill na’s (e.g. NaN’s) with ”. inplace is possible but should be avoided as it will be deprecated: df.fillna(”, inplace=True) To fill only a single column: df.column1 = df.column1.fillna(”) One can use df[‘column1’] instead of df.column1.
According to the IEEE standard, NaN values have the odd property that comparisons involving them are always false. That is, for a float f, f != f will be true only if f is NaN. Note that, as some comments below have pointed out, not all compilers respect this when optimizing code. For any compiler … Read more
Try this code: isNaN(parseFloat(“geoff”)) For checking whether any value is NaN, instead of just numbers, see here: How do you test for NaN in Javascript?
I believe DataFrame.fillna() will do this for you. Link to Docs for a dataframe and for a Series. Example: In [7]: df Out[7]: 0 1 0 NaN NaN 1 -0.494375 0.570994 2 NaN NaN 3 1.876360 -0.229738 4 NaN NaN In [8]: df.fillna(0) Out[8]: 0 1 0 0.000000 0.000000 1 -0.494375 0.570994 2 0.000000 0.000000 … Read more
jwilner’s response is spot on. I was exploring to see if there’s a faster option, since in my experience, summing flat arrays is (strangely) faster than counting. This code seems faster: df.isnull().values.any() import numpy as np import pandas as pd import perfplot def setup(n): df = pd.DataFrame(np.random.randn(n)) df[df > 0.9] = np.nan return df def … Read more
RukTech’s answer, df.set_value(‘C’, ‘x’, 10), is far and away faster than the options I’ve suggested below. However, it has been slated for deprecation. Going forward, the recommended method is .iat/.at. Why df.xs(‘C’)[‘x’]=10 does not work: df.xs(‘C’) by default, returns a new dataframe with a copy of the data, so df.xs(‘C’)[‘x’]=10 modifies this new dataframe only. … Read more
Don’t drop, just take the rows where EPS is not NA: df = df[df[‘EPS’].notna()]