[英]Making a list from a list of lists, using the same repeated index number for each list created by the enumerate method
我的 function desired_headers() 从“结果” object 中删除元组,其中元组中的第一个元素与标题列表中的任何字符串都不匹配。 ['Fish', 'Dolphin']
from result1 object & ['Local Auth', 'bucket']
已从 result3 中删除。
然后它打印索引号和元组列表,如下面的当前 output 所示。
我的目标是使用我当前的 output 将元组列表“重新打包”并存储在 object 中。
headers2 = ['Group', 'Owner', 'Person in charge', 'Type of Service',
'Registered Care Categories*', 'Specialist Care Categories',
'Languages Spoken by Staff (other than English)','Single Rooms',
'Shared Rooms','Facilities & Service']
result1 = [['Group', 'MacIntyre'], ['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'], ['Fish', 'Dolphin'], ['Shared Rooms', '4']]
result2 = [['Group', 'Jameseson'], ['Type of Service', 'bad'], ['Shared Rooms', '8']]
result3 = [['Group', 'MacIntyre'], ['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'], ['Shared Rooms', '4'], ['Local Auth', 'bucket']]
results = [result1, result2, result3]
#Removes tuples which the first element in the tuple does not matcch any string in headers2 list
def desired_headers():
for index, z in enumerate(list(range(len(results)))):
for i in list(range(len(results[z]))):
if any(x in headers2 for x in results[z][i]):
print(index, results[z][i])
desired_headers()
当前 Output:
0 ['Group', 'MacIntyre']
0 ['Person in charge', ' Vivienne Donald (Manager)']
0 ['Type of Service', 'good']
0 ['Shared Rooms', '4']
1 ['Group', 'Jameseson']
1 ['Type of Service', 'bad']
1 ['Shared Rooms', '8']
2 ['Group', 'MacIntyre']
2 ['Person in charge', ' Vivienne Donald (Manager)']
2 ['Type of Service', 'good']
2 ['Shared Rooms', '4']
所需的 Output:
[[['Group', 'MacIntyre'],
['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'],
['Shared Rooms', '4']],
[['Group', 'Jameseson'], ['Type of Service', 'bad'], ['Shared Rooms', '8']],
[['Group', 'MacIntyre'],
['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'],
['Shared Rooms', '4']]]
你需要这样的东西吗? 这会将所有内容分组在一个列表中
# Removes tuples which the first element in the tuple does not matcch any string in headers2 list
def desired_headers():
grouped_elements = []
for index, z in enumerate(list(range(len(results)))):
inner_loop = []
for i in list(range(len(results[z]))):
if any(x in headers2 for x in results[z][i]):
inner_loop.append(results[z][i])
grouped_elements.append(inner_loop)
print(grouped_elements)
尝试这个
lv1 = [] # initialize an empty list which we will append to based on our criteria
for lst in results: # loop though each element of the list, which itself is a list
lv2 = []
for el in lst: # Same as above
if el[0] in headers2: # check if first element of the list exists in headers2
lv2.append(el)
lv1.append(lv2)
lv1
或者如果你想要一个 function
def desired_headers(list_of_lists, inclution_list):
lv1 = []
for lst in list_of_lists:
lv2 = []
for el in lst:
if el[0] in inclution_list:
lv2.append(el)
lv1.append(lv2)
return lv1
desired_headers(list_of_lists=[result1, result2, result3], inclution_list=headers2)
或者,如果您熟悉列表理解
result = []
for lst in [result1, result2, result3]:
result.append([el for el in lst if el[0] in headers2])
result
您可以通过列表理解来做到这一点:
filtered = [ [sl for sl in r if sl[0] in headers2] for r in results ]
output:
print(filtered)
[
[
['Group', 'MacIntyre'],
['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'],
['Shared Rooms', '4']
],
[
['Group', 'Jameseson'],
['Type of Service', 'bad'],
['Shared Rooms', '8']
],
[
['Group', 'MacIntyre'],
['Person in charge', ' Vivienne Donald (Manager)'],
['Type of Service', 'good'],
['Shared Rooms', '4']
]
]
如果您有大量数据,您可能希望使用headers2
来制作一组用于列表理解
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