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x = ["python", "java", "c++", "c#", "javascript"] | |
word1 = None | |
max_length = 0 | |
for i in x: | |
if len(i) > max_length: | |
max_length = len(i) | |
word1 = i | |
print(word1) | |
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# Расчёт квадрата | |
# На вход программе поступает список чисел. | |
# Каждый элемент — сторона квадрата. | |
# Нужно создать новый список, где каждому исходному элементу соответствует площадь квадрата: | |
# Преимущества перебора по элементам списка, а не по индексам | |
# Если нам нужно измнить исходных список, то лучше переберать по элементам |
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# Конечно, лаконичнее выглядит цикл, который перебирает сами элементы списка, | |
# а не его индексы. Но у перебора индексов есть большое преимущество. | |
# Такой цикл может изменять значения элементов. | |
# Допустим, список хранит цены товаров. Задача: уменьшить каждую цену на 1000: | |
prices = [1500, 2999, 7499, 3220] | |
for i in range(len(prices)): | |
prices[i] -= 1000 | |
print('Результат:', prices) |
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numbers = [1,2,3,4,5] # Список в который нужно добавить 6, 7 ... в конец списка с помощью insert | |
len_numbers = len(numbers) # Считаем колличество элементов в списке и присваиваем переменной | |
numbers.insert(len_numbers, 6) | |
len_numbers = len(numbers) | |
numbers.insert(len_numbers, 7) | |
print(numbers) |
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# Задача Яндекс.Практикум, факультет Data Science | |
# В списке yellow_submarine_chorus некоторые значения повторяются. | |
# Создайте список yellow_submarine_briefly, в котором каждый элемент исходного списка встречается только раз. | |
yellow_submarine_chorus = ['we', 'all', 'live', 'in', 'a', 'yellow', 'submarine', 'yellow', 'submarine', 'yellow', 'submarine', 'we', 'all', 'live', 'in', 'a', 'yellow', 'submarine', 'yellow', 'submarine', 'yellow', 'submarine'] | |
yellow_submarine_briefly = [] | |
# ваш код здесь | |
for i in yellow_submarine_chorus: |
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country | capital | |
---|---|---|
Abkhazia | Sukhumi | |
Afghanistan | Kabul | |
Akrotiri and Dhekelia | Episkopi Cantonment | |
Albania | Tirana | |
Algeria | Algiers | |
American Samoa | Pago Pago | |
Andorra | Andorra la Vella | |
Angola | Luanda | |
Anguilla | The Valley |
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def anagram(word1, word2): # Передаем 2 слова, которые хотим сравнить на анаграммы | |
word1 = word1.lower() # Приводим все к нижнему регистру в первом слове | |
word2 = word2.lower() # Приводим все к нижнему регистру во втором слове | |
word1 = ''.join(sorted(word1)) # Сортируем все буквы в слове по алфавиту в первом слове | |
word2 = ''.join(sorted(word2)) # Сортируем все буквы в слове по алфавиту во втором слове | |
if word1 == word2: # Сравниваем отсортированные слова | |
print('Анаграммы') # Если отсортированные слова равны, то выводим в консоль сообщение об анаграмме |
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function palindrome(word) { | |
if (word === word.split('').reverse().join('')) { | |
return true; | |
} else { | |
return false; | |
} | |
} |
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function count_letters(word) { | |
var count = 0; | |
for (var i = 0; i < word.length; i++) { | |
count += 1; | |
} | |
return count; | |
} |