#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’]
my_list = ['aabbcc', 'a1b1c1']
string = 'curl -X DELETE https://marina.ring.com/'
my_new_list = [string + x for x in my_list]
print (my_new_list)
#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}
word = "abracadabra"
dict_last_occ = {}
for letter in word:
dict_last_occ[letter] = word.rfind(letter)
print(dict_last_occ)
#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}
list_initial = [1, 2, 4, 6, 2, 7, 1, 2, 1, 2, 2]
dict_freq = {}
for number in list_initial:
if number not in dict_freq.keys():
dict_freq[number] = 1
else:
dict_freq[number] += 1
print(dict_freq)
#3. There is a list of integers, return count of even elements:
[1, 2, 3, 4, 5, 6, 6, 6,7, 8, 22] -> 7
my_list = [1, 2, 3, 4, 5, 6, 6, 6,7, 8, 22]
even_count = 0
for num in my_list:
if num % 2 == 0:
even_count += 1
print("Even numbers in the list: ", even_count)
#4. 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]
list_initial = [0, 3, 3, 6, 12, 7, 4, 21]
list_new = [list_initial[0]]
for number in list_initial [1:]:
if number % list_initial.index(number)==0:
list_new.append(number)
print(list_new)
#але тут у мене 3 повертає двічі. мабуть, обидва рази трійка ділиться на один індекс. пробую з каунтером підкоригувати - поки не виходить
#5. 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'
s = "quick BROWN fox JUMPS over THE lazy DOG"
l = list()
for i, c in enumerate(s):
if i % 2 == 0:
l.append(c.upper())
else:
l.append(c.lower())
print("".join(l))