如何在python3中创建字典？

我有一个数组，如何在python3中创建字典？

months=['January','February','March','April','May','June','July','August','September','October','November','December']

，我希望把它作为字典一样

months = {1:'January', 2:'February',....}

我试图enumerate()但没有任何运气。如何将序列号分配给列表以制作字典？在此先感谢！

来源

2016-06-09 jaykodeveloper

如果你想坚持枚举，你可以做这样的：

months =['January','February','March','April','May','June','July','August','September','October','November','December'] 

d_months = {} 

for i, month in enumerate(months): 
    d_months[i+1] = month 

print(d_months)

输出：

{1: 'January', 2: 'February', 3: 'March', 4: 'April', 5: 'May', 6: 'June', 7: 'July', 8: 'August', 9: 'September', 10: 'October', 11: 'November', 12: 'December'}

由于枚举从0开始，你需要使用[i+1]

来源

2016-06-09 02:52:43

谢谢！它运作良好！ – jaykodeveloper

试试这个。

months= 
['January','February','March','April','May','June','July','August','September','October','November','December'] 

dict = {i+1:months[i] for i in range(12)}

正如评论中指出的那样，使用范围（len（months））。由于列表的长度在这种情况下是已知的，我用12

来源

2016-06-09 02:50:35

应该n硬编码数组大小 - 使用'len（月）' –

@IvanNevostruev是。应该提到这一点。由于我们在这种情况下事先知道了这个长度，所以我没有看到需要。将编辑我的答案:) –

@AbhirathMahipal我也使用'范围'。谢谢！ – jaykodeveloper

尝试使用eumerate这样，迫使它开始与1

months =['January','February','March','April','May','June','July','August','September','October','November','December'] 
months = {i:m for i,m in enumerate(months,1)} 
print months

会产生

{1: 'January', 2: 'February', 3: 'March', 4: 'April', 5: 'May', 6: 'June', 7: 'July', 8: 'August', 9: 'September', 10: 'October', 11: 'November', 12: 'December'}

来源

2016-06-09 02:59:37 nehemiah

由于dict()构造直接从键值对建立一个字典，这也适用：

dict(enumerate(months, 1))

这里有一些测试，尽管这ISN”对您的使用情况非常重要：

timeit dict(enumerate(months, 1)) 
The slowest run took 10.34 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.37 µs per loop 

timeit {i:m for i,m in enumerate(months,1)} 
The slowest run took 7.77 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.28 µs per loop 

timeit {i+1:months[i] for i in range(12)} 
The slowest run took 5.42 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.89 µs per loop

来源

2016-06-09 04:19:11

感谢您分享比较。你写了那个代码吗？ – jaykodeveloper

如何在python3中创建字典？

回答

相关问题