本文是學習齊偉老師的《python全棧工程師》課程的筆記,歡迎學習交流。同時感謝齊老師的精彩傳授!
習題01:
将字元串“python”轉化為清單(記為lst),然後完成如下操作:
– 将字元串 ‘rust’ 中的每個字母作為獨立元素追加到lst中
– 對 lst 排序
– 删除 lst 中的重複元素
s = 'python'
lst = list(s)
r = 'rust'
lst.extend(r)
lst.sort()
list(set(lst))
運作效果圖:

習題02
編寫程式,實作如下功能:
– 使用者輸入國家名稱
– 列印出所輸入國家名稱及其首都
nations = {'China': 'Beijing', 'Japan': 'Tokyo', 'India': 'New Delhi', 'Sweden': 'Stockholm', 'Russian': 'Moscow', 'Germany': 'Berlin', 'UK': 'London', 'French': 'Paris', 'Swiss': 'Bern', 'Egypt':'Cairo', 'Australia': 'Canberra', 'New Zealand': 'Wellington', 'Canada': 'Ottawa', 'USA': 'Washington', 'Cuba': 'Havana', 'Brazil': 'Brasilia'}
name = input('input a name of country:')
capital = nations.get(name)
print('the country:', name)
print('the capital:', capital)
運作效果圖:
習題03
有如下技術棧名稱集合:skills = {‘Python’, ‘R’, ‘SQL’, ‘Git’, ‘Tableau’, ‘SAS’ }。假設自己的技術是:
myskills = {‘Python’, ‘R’},判斷自己所掌握的技術是否在上述技術棧範圍之内。
skills = {'Python', 'R', 'SQL', 'Git', 'Tableau', 'SAS' }
myskills = {'Python', 'Java'}
myskills.intersection(skills)
運作效果圖:
習題04
找出以下兩個字典共有的鍵:
{’a’: 1, ‘b’: 2, ‘c’: 3, ‘d’: 4}
{‘b’: 22, ‘d’: 44, ‘e’: 55, ‘f’: 77}
d1 = {’a': 1, 'b': 2, 'c': 3, 'd': 4}
d2 = {'b': 22, 'd': 44, 'e': 55, 'f': 77}
# 方法一:
d1_set = set(d1.keys())
d2_set = set(d2.keys())
common_key = d1_set.intersection(d2_set)
# 方法二:适用于python3.6+的版本
d1.keys() & d2.keys()
運作效果圖:
習題05
字元串:songs = ‘You raise my up so I can stand on mountains You raise my up to walk on stormy seas I am strong when I am on our shoulders You raise me up to mroe than I can be’
制作上述字元串的單詞表,統計每個單詞的出現次數
小編參考:
songs = 'You raise my up so I can stand on mountains You raise my up to walk on stormy seas I am strong when I am on our shoulders You raise me up to mroe than I can be'
lst = songs.split(' ')
new_lst = {}
for i in range(len(lst)):
if lst[i] in new_lst:
new_lst[lst[i]] += 1
else:
new_lst[lst[i]] = 1
print(new_lst)
運作效果圖: