Question

假设我创建了一个字典并存储了以下键和值：

bod = {“Test1”：12345，“Test2”：1323242，...}

现在，如果我创建一个新列表并具有以下值

bof = [“Test3”，“Test1”，“Test4”，“Test2”]

是否可以使用dict变量作为调用将其与列表匹配，并使用以下代码（伪）

在新变量内分配匹配键的值

for l in bof:
    newbof = line.split()
      try:
        newvalues = bod[values]
        print newvalues

Answer 1

>>> bod = { "Test1" : 12345, "Test2" : 1323242 }
>>> bof = ["Test3", "Test1", "Test4", "Test2"]
>>> [bod.get(x) for x in bof]
[None, 12345, None, 1323242]

其他变体

>>> [bod.get(x, 0) for x in bof]
[0, 12345, 0, 1323242]
>>> [bod[x] for x in bof if x in bod]
[12345, 1323242]

Answer 2

根据我的理解，无法完全理解您的问题，

newvalues=[]
for i in bof:
  if bod.get(i):
    newvalues.append(bod[i])
print newvalues

优化

newvalues=[bod[i] for i in bof if bod.get(i)]

Answer 3

据我所知：

In [325]: bod = { "Test1" : 12345, "Test2" : 1323242}

In [326]: bof = ["Test3", "Test1", "Test4", "Test2"]

In [327]: [bod[i ]for i in bof if i in bod]
Out[327]: [12345, 1323242]

或者如果你想保留不在bod中的值：

In [332]: [bod[x] if x in bod else x for x in bof]
Out[332]: ['Test3', 12345, 'Test4', 1323242]

或

[bod.get(x,x) for x in bof]
Out[333]: ['Test3', 12345, 'Test4', 1323242]

另请注意，虽然使用get更简洁，但使用in会更快：

In [337]: % timeit [bod[x] if x in bod else x for x in bof]
1000000 loops, best of 3: 1.39 us per loop

In [338]: % timeit [bod.get(x,x) for x in bof]
100000 loops, best of 3: 1.88 us per loop

使用字典作为调用

3 个答案: