Storing values for each iteration Numpy Python

user_15217016 0
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The code down below goes through the Vals function and it assorts through the Numbers value and adds up all the sums after the sorting. I am trying to append the SUM values to the T_SUM to store the values of each SUM for each sort.

Vals= np.arange(start=60, stop=105, step=5)
Numbers = np.array([123.6,       130 ,       150,        110.3748,     111.6992976,
 102.3165566,   97.81462811 , 89.50038472 , 96.48141473 , 90.49956702, 65])
T_Sum = np.array([])
p= 0 

while len(Vals) != p:
    Numbers= np.where(Numbers >= Vals[p],Numbers ,0 )
    p = p + 1
    SUM = np.sum(Numbers)
    T_Sum = np.concatenate((T_Sum, SUM))

print(T_Sum)
while-loop numpy sum arrays python
2 Answers

You can use masked arrays to mask values you desire and then sum them up all in one line:

T_Sum = np.ma.masked_array(np.repeat(Numbers[None,:],Vals.size,0),mask=[Numbers<Vals[:,None]]).sum(-1).data

output:

array([1167.28664878, 1167.28664878, 1102.28664878, 1102.28664878,
       1102.28664878, 1102.28664878, 1012.78626406,  922.28669704,
        727.9906542 ])
user_49759810

This should do the same thing:

Vals= np.arange(start=60, stop=105, step=5)
Numbers = np.array([123.6,       130 ,       150,        110.3748,     111.6992976,
 102.3165566,   97.81462811 , 89.50038472 , 96.48141473 , 90.49956702, 65])
T_sum = np.empty(len(Vals))
count = 0

for i in Vals:
    Numbers= np.where(Numbers >= Vals[p],Numbers ,0 )
    SUM = np.sum(Numbers)
    T_sum[count] = SUM
    count += 1
user_152702870
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