如何在python和Robot框架中对字典进行切片

Question

切片可用于python 中的列表

list1 =[1,2,3,4,5,6]
list1[:3]
[1, 2, 3]

类似地，切片或者类似于dictionary的东西？

dict1 = {1":a",2:"b",3:"c",4:"d",5:"e"} 

我想得到字典的任何3个(可以是随机的(元素，只需提供数字(如上面为列表[:2]提供的(，那么我应该得到字典以下的元素

dict1 = {1":a",2:"b"} # After slicing

如何在CCD_ 2&Robot-framework？

Answer 1

也许这是一个您可以考虑的解决方案，因为dict不能作为list:访问

dict1 = {1:"a",2:"b",3:"c",4:"d",5:"e"}
def take(dct, high=None, low=None):
return dict(list(dct.items())[low:high])
print(take(dict1, 3)) #=> {1: 'a', 2: 'b', 3: 'c'}
print(take(dict1, 5, 2)) #=> {3: 'c', 4: 'd', 5: 'e'}

Answer 2

仅使用Robot Framework关键字提供两个备选方案。从本质上讲，他们遵循类似的方法。从字典中获取密钥，然后对其进行切片，然后以所需格式修改或重新创建字典。

除非有特定的原因不想使用Python，否则我认为这个功能应该由Python关键字而不是Robot Framework提供。

*** Settings ***
Library    Collections
*** Variables ***
&{dict1}    1=a    2=b    3=c    4=d    5=e
&{dict2}    1=a    2=b    3=c    4=d    5=e
&{result}   3=c    4=d    5=e 
*** Test Cases ***
TC - keep items 3, 4 & 5
# Keey
Keep Slice In Dictionary    ${dict1}    ${5}    ${2}
Log Many    ${dict1}
Dictionaries Should Be Equal    ${dict1}    ${result}    
${slice}       Get Slice From Dictionary    ${dict2}    ${5}    ${2}
Log Many    ${slice}
Dictionaries Should Be Equal    ${dict1}    ${slice}
*** Keywords ***
Keep Slice In Dictionary
[Documentation]
...    Modifies the dictionary to only leave the slice.
...    
...    The keys of the dictionary are converted into a list. Then
...    this list is spliced. This list is then used to filter out
...    the unwanted keys.
...    
...    Note: this keyword modifies the provided dictionary.
...    
...    Arguments:
...    - dict    (dictionary)    The dictionary that needs to be modified
...    - high    (integer)       The last item to be kept.
...    - low     (integer)       The first item of the slice. (defaults to 0)
...    
...    Returns:    None          Modifies the provided dictionary.
...    
[Arguments]    ${dict}    ${high}    ${low}=${0}
${keys_list}        Get Dictionary Keys    ${dict}
${filtered_keys}    Get Slice From List    ${keys_list}    ${low}    ${high}
Keep In Dictionary     ${dict}    @{filtered_keys}
Get Slice From Dictionary
[Documentation]
...    Get a slice of sequential keys from a dictionary
...    
...    The keys of the dictionary are converted into a list. Then
...    this list is spliced. This list is then used to create a new
...    Dictionary with the filtered keys.
...    
...    Arguments:
...    - dict    (dictionary)    The source dictionary
...    - high    (integer)       The last item to be kept.
...    - low     (integer)       The first item of the slice. (defaults to 0)
...    
...    Returns:  (dictionary     A dictionary with the desired keys.
...    
[Arguments]    ${dict}    ${high}    ${low}=${0}
${keys_list}        Get Dictionary Keys    ${dict}
${filtered_keys}    Get Slice From List    ${keys_list}    ${low}    ${high}
${return_dict}    Create Dictionary
:FOR    ${item}    IN    @{filtered_keys}
        Set To Dictionary    ${return_dict}   ${item}    ${dict['${item}']}
[Return]     ${return_dict}

Answer 3

我想获得字典的任意3个元素(可以是随机的(

不需要构造所有字典项的列表。您可以使用dict.items和itertools.islice对固定数量的项目进行切片：

from itertools import islice
def get_n_items(d, n):
return dict(islice(d.items(), 0, n))
dict1 = {1:"a", 2:"b", 3:"c", 4:"d", 5:"e"} 
get_n_items(dict1, 2)  # {1: 'a', 2: 'b'}
get_n_items(dict1, 3)  # {1: 'a', 2: 'b', 3: 'c'}

对于Python 3.6+，作为CPython 3.6和3.7+中的一个实现细节，这相当于按插入顺序获取第一个n项。