numpy count the number of 1s in array
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))
{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}numpy count the number of 1s in array
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))
{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}python 3 numbers of a range is even
lst = []
for i in range(1000,3001):
flag = 1
for j in str(i): # every integer number i is converted into string
if ord(j)%2 != 0: # ord returns ASCII value and j is every digit of i
flag = 0 # flag becomes zero if any odd digit found
if flag == 1:
lst.append(str(i)) # i is stored in list as string
print(",".join(lst))Copyright © 2021 Codeinu
Forgot your account's password or having trouble logging into your Account? Don't worry, we'll help you to get back your account. Enter your email address and we'll send you a recovery link to reset your password. If you are experiencing problems resetting your password contact us
