I don't think that's the case. I believe OP is saying that doing[float] * [Int64]the dtype gets coerced toFloat64, but containing the valuenp.nan(rather than coercing to the expectedpd.NA). With this, doing the sum then gives the valuenp.nanrather than the expected0. ...
want to calculate the sum of the numbers appear in an array that is mix of NaNs and numbersIt is not trivial actually. Look at the following and let me know if there is anything unclear. Also, I let you check that it manages well limit cases, and...
【题目】18. I have decided to go on a trip to Hu nan Islind this sumtmer. (对画线部分提问)odo t
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07.Sum of numbers from 0 to N 描述: 我们要生成一个函数,该函数计算从0开始的序列,并结束,直到序列之后的给定数字: 0 1 3 6 10 15 21 28 36 45 55 ... 由...创建 0 + 0 + 1 + 2 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 4 + 0 + 1 + 2 + 3 + 4 + 4 + 5 + 6,...
Why sum is getting nan as the output? #include <stdio.h> int main(void){ int input,j; float sum; for(j = 1; j<=2;j++){ printf("Enter the mark for module %d ",j); scanf("%d",&input); sum += input; } printf("The Sum is: %f",sum); return 0; } Inputs: 1,2 Ac...
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2.0 2 NaN NaN 3 3.0 3.0 df1.sum(axis=1, skipna=False) Out[6]: 0 NaN 1 4.0 2 NaN 3 6.0 dtype: float64 df1.sum(axis=1, skipna=True) Out[7]: 0 1.0 1 4.0 2 0.0 3 6.0 dtype: float64 df1.sum(axis=1, min_count=1) Out[7]: 0 1.0 1 4.0 2 NaN 3 6.0 dtype: float...
I want to sum the columns except the first two and then replace the values not NaN with the sum the column results: the result should like this: data = {'Org': ['Tom', 'Kelly', 'Rick', 'Dave','Sara','Liz'], 'A': ['NaN', 1, 1, 1, 'NaN', 'NaN'], 'B': [5, 5,...