python字典fromkeys()方法中的坑
阿新 • • 發佈:2019-03-21
form tip 本質 pre 聯動 字典 range end pen
tips: fromkeys方法會返回一個新的字典,對原字典無影響
自定操作中的fromkeys()方法接收兩個參數,第一個參數為一個可叠代對象,作為返回字典的key,第二個參數為value,默認為None,具體用法如下:
li = [1,2,3]
dic1 = dict.fromkeys(li)
dic2 = dict.fromkeys(li,[])
print(dic1) # {1: None, 2: None, 3: None}
print(dic2) # {1: [], 2: [], 3: []}
此時我為dic2中key為1的列表增加一個元素‘test’,如下:
dic2[1].append(‘test‘) print(dic2) # {1: [‘test‘], 2: [‘test‘], 3: [‘test‘]}
竟然把三個列表的值都給改了,這是為啥呢?先打印下他們的內存地址
print("dic2[1]:{}\ndic2[2]:{}\ndic2[3]:{}".format(id(dic2[1]),id(dic2[2]),id(dic2[3])))
# dic2[1]:1714986428808
# dic2[2]:1714986428808
# dic2[3]:1714986428808
原來它的所有鍵都指向了同一個內存地址,這也就不難怪修改其中一個而引發聯動了,因為本質上只有一個列表。因此,在字典中定義不同的列表不要用fromkeys方法,還是老老實實定義吧
dic2 = {1: [], 2: [], 3: []} print(id(dic2[1])) # 1657985662344 print(id(dic2[2])) # 1657986500680 print(id(dic2[3])) # 1657986501960 dic2[1].append(‘test‘) print(dic2) # {1: [‘test‘], 2: [], 3: []}
使用循環來產生多key的字典:
dic2 = {}
for k in range(10):
dic2[k] = []
print(dic2) # {0: [], 1: [], 2: [], 3: [], 4: [], 5: [], 6: [], 7: [], 8: [], 9: []}
dic1 = {1:2} dic2 = dic1.fromkeys([1,2,3],‘test‘) print(dic1) # {1: 2} print(dic2) # {1: ‘test‘, 2: ‘test‘, 3: ‘test‘}
python字典fromkeys()方法中的坑