python基础_字典

Posted 2020-12-11 逐梦无惧_数据分析

empty_dict = {}
a_dict = {‘one‘:1,‘two‘:2,‘three‘:3}
print("{}".format(a_dict))
print("{}".format(len(a_dict)))
another_dict = {‘x‘:‘printer‘,‘y‘:5,‘z‘:[‘star‘,‘circle‘,9]}
print("{!s}".format(len(another_dict)))
print("{}".format(another_dict[‘z‘]))
a_new_dict = a_dict.copy()
print("{}".format(a_new_dict))
print("{}".format(a_dict.keys()))
print("{}".format(a_dict.values()))
print("{}".format(a_dict.items()))
if ‘c‘ not in another_dict:
print("{}".format(another_dict.keys()))
print("{}".format(a_dict.get(‘three‘)))
print("{}".format(a_dict.get(‘four‘)))
print("{}".format(a_dict.get(‘four‘,‘not in dict‘)))

{‘one‘: 1, ‘two‘: 2, ‘three‘: 3}
3
3
[‘star‘, ‘circle‘, 9]
{‘one‘: 1, ‘two‘: 2, ‘three‘: 3}
dict_keys([‘one‘, ‘two‘, ‘three‘])
dict_values([1, 2, 3])
dict_items([(‘one‘, 1), (‘two‘, 2), (‘three‘, 3)])
dict_keys([‘x‘, ‘y‘, ‘z‘])
3
None
not in dict

 

a_dict = {‘one‘:1,‘two‘:2,‘three‘:3}
dict_copy = a_dict.copy()
ordered_dict1 = sorted(dict_copy.items(),key=lambda item:item[0])
print("{}".format(ordered_dict1))
ordered_dict2 = sorted(dict_copy.items(),key=lambda item:item[1])
print("{}".format(ordered_dict2))
ordered_dict3 = sorted(dict_copy.items(),key=lambda x:x[1],reverse=True)
print("{}".format(ordered_dict3))

[(‘one‘, 1), (‘three‘, 3), (‘two‘, 2)]
[(‘one‘, 1), (‘two‘, 2), (‘three‘, 3)]
[(‘three‘, 3), (‘two‘, 2), (‘one‘, 1)]