Sum of elements within an array within list

Question

I'm trying to calculate the np.sum of the elements within each array. I tried in stead of np.sum(outcome_list[0] == 'H' to just leave it as np.sum(outcome_list[j] == 'H' so that each "list" would have its own data set on the total number of heads, but it didn't like it. The bigger question is, how would I construct an array with a given base list and the action to be done in each element of that list?

EDIT:

the throw_a_coin definition

def throw_a_coin(N):
    return np.random.choice(['H','T'], size=N)
N =40

trials (as shown above) is the set to be acted upon

for i in trials:
    throws = throw_a_coin(i)
    outcome_list.append(throws)

for j in outcome_list:
    print("Number of Heads:", np.sum(outcome_list[0] == 'H'))
    print (j)

EDIT 2:

problem resolved with the one shown below, however I'm getting more than 13 numbers for "probabilities" - it seems that the system is running through the trials list more than once.

def throw_a_coin(N):
    return np.random.choice(['H','T'], size=N)

trials = [10, 30, 50, 70, 100, 130, 170, 200, 500, 1000, 2000, 5000, 10000]

for i in trials:
    throws = throw_a_coin(i)
    outcome_list.append(throws)

probabilities = []

for j in outcome_list:
    print("Number of Heads:", np.sum(j == 'H'))
    print("Number of Throws:", len(j))
    print("p = Number of Heads/Total Throws:", (np.sum(j == 'H'))/len(j))
    probabilities.append((np.sum(j =='H'))/len(j))
    print (j)
    print("\n")

print(probabilities)

Can you please attach the code instead of the photo of the code? — Sohum Sachdev
– Sohum Sachdev, Commented Sep 3, 2017 at 5:08
@Rishav - yes, count the number of heads per each of the trials — aiwan
– aiwan, Commented Sep 3, 2017 at 5:16

Sohum Sachdev · Accepted Answer · 2017-09-03 05:21:22Z

1

You were nearly there! You just needed to replace

print("Number of Heads:", np.sum(outcome_list[0] == 'H'))

with

print("Number of Heads:", np.sum(j == 'H'))

Here is the complete answer:

trials = [10, 30, 50, 70, 100, 130, 170, 200, 500, 1000, 2000, 5000, 10000]

N =40
def throw_a_coin(N):
    return np.random.choice(['H','T'], size=N)

outcome_list = []
for i in trials:
    throws = throw_a_coin(i)
    outcome_list.append(throws)

for j in outcome_list:
    print("Number of Heads:", np.sum(j == 'H'))
    print (j)

answered Sep 3, 2017 at 5:21

Sohum Sachdev

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1 Comment

aiwan Over a year ago

I've also added "EDIT 2" to the question - it seems that it's running through "trials" more than once?

Jeffrey Ram · Accepted Answer · 2017-09-03 05:26:24Z

1

Try this:

print("Number of Heads:", (j == 'H').sum())

edited Sep 3, 2017 at 5:26

answered Sep 3, 2017 at 5:20

Jeffrey Ram

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Comments

Sam Chats · Accepted Answer · 2017-09-03 05:20:34Z

0

If output_list would be a numpy array and if each row in the matrix represent each trial, you can get the sum in an efficient way as follows:

head_sum = output_list.sum(axis=1)

answered Sep 3, 2017 at 5:20

Sam Chats

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Comments

xrisk · Accepted Answer · 2017-09-03 05:22:12Z

0

I would use a list comprehension followed by a len call.

for j in outcome_list:
    print("Number of Heads:", len([x for x in j if x == 'H'])
    print (j)

DISCLAIMER: I have zero experience with Numpy. However, this would be the general Pythonic way to do it without resorting to count.

answered Sep 3, 2017 at 5:22

xrisk

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1 Comment

Patrick Haugh Over a year ago

This is pretty good, but I would recommend sum(1 for x in j if x == 'H') By using a generator comprehension, we avoid actually building the list of Hs. It's a tiny difference, but I just thought I would mention it.

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