There is list of mac addresses create list of commands to delete this

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# 1. There is a list of mac addresses - create a list of commands to delete this mac
# [‘aabbcc’, ‘a1b1c1’] -> [‘curl -X DELETE https://marina.ring.com/aabbcc’, ‘curl -X DELETE https://marina.ring.com/a1b1c1’]
mac = ['aabbcc', 'a1b1c1']
curl = 'curl -X DELETE http://sds.com/'
link =[]
for i in mac:
link.append([curl+i])
print(link)
# 2. There is a string, for each letter return the position of last occurrence.
# ‘abracadabra’ -> {'a': 10, 'b': 8, 'r': 9, 'c': 4, 'd': 6}
st = 'abracadabra'
a = {k:v for v, k in enumerate(st)}
print(a)
# 2.*There is a list of integers, create a dict with count of each element (have a look at defaultdict)
# [1, 2, 4, 6, 2, 7, 1, 2, 1, 2, 2] -> {1: 3, 2:5, 4:1, 6:1, 7:1}
#
from collections import defaultdict
st = [1, 2, 4, 6, 2, 7, 1, 2, 1, 2, 2]
result = defaultdict(int)
for integer in st:
result[integer] += 1
print(result.items())
# There is a list of integers, return count of even elements:
# [1, 2, 3, 4, 5, 6, 6, 6,7, 8, 22] -> 7
data = [1, 2, 3, 4, 5, 6, 6, 6, 7, 8, 22]
result = []
for elem in data:
if elem % 2 == 0:
result.append(elem)
print(len(result))
# There is a list of integers, return only those elements that can be divided by position index (add 0s)
# [0, 3, 3, 6, 12, 7, 4, 21] -> [0, 3, 6, 12, 21]
data = [0, 3, 3, 6, 12, 7, 4, 21]
result = [v for k, v in enumerate(data) if k == 0 or v % k == 0]
print(result)
# Turn letters on even positions uppercase and on odd positions lowercase
# "quick BROWN fox JUMPS over THE lazy DOG" -> 'QuIcK BrOwN FoX JuMpS OvEr tHe lAzY DoG'
sent = 'quick BROWN fox JUMPS over THE lazy DOG'
print(sent)
print(''.join([v.upper() if k % 2 == 0 else v.lower() for k, v in enumerate(sent)]))