Python 从元组列表中获取元组的前几个元素
我有一个元组列表的结构,如:-Python 从元组列表中获取元组的前几个元素,python,python-3.x,Python,Python 3.x,我有一个元组列表的结构,如:- [[(1, 1, 96), (1, 2, 95), (0, 5, 23), (0, 6, 22)], [(2, 1, 145), (1, 2, 144), (10, 3, 143), (2, 4, 142)]] 我基本上想从中获得2个元组列表。一个元组用于前两列,另一个元组用于第三列。 期望输出:- [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3
[[(1, 1, 96),
(1, 2, 95),
(0, 5, 23),
(0, 6, 22)],
[(2, 1, 145),
(1, 2, 144),
(10, 3, 143),
(2, 4, 142)]]
我基本上想从中获得2个元组列表。一个元组用于前两列,另一个元组用于第三列。
期望输出:-
[[(1, 1),
(1, 2),
(0, 5),
(0, 6)],
[(2, 1),
(1, 2),
(10, 3),
(2, 4)]]
&&
如何在python中实现这一点
[[(a, b) for a, b, *c in r] for r in arr]
# => [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3), (2, 4)]]
[[tuple(c) for a, b, *c in r] for r in arr]
# => [[(96,), (95,), (23,), (22,)], [(145,), (144,), (143,), (142,)]]
针对评论:
def slice_nested_array(arr, start, stop=None, step=1):
if stop is None:
stop = len(arr[0][0])
return [[tuple(l[start:stop:step]) for l in r] for r in arr]
slice_nested_array(arr, 0, 2)
# => [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3), (2, 4)]]
slice_nested_array(arr, 2)
# => [[(96,), (95,), (23,), (22,)], [(145,), (144,), (143,), (142,)]]
针对评论:
def slice_nested_array(arr, start, stop=None, step=1):
if stop is None:
stop = len(arr[0][0])
return [[tuple(l[start:stop:step]) for l in r] for r in arr]
slice_nested_array(arr, 0, 2)
# => [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3), (2, 4)]]
slice_nested_array(arr, 2)
# => [[(96,), (95,), (23,), (22,)], [(145,), (144,), (143,), (142,)]]
只需保持简单,循环嵌套列表,并获取所需内容:
lst = [[(1, 1, 96),
(1, 2, 95),
(0, 5, 23),
(0, 6, 22)],
[(2, 1, 145),
(1, 2, 144),
(10, 3, 143),
(2, 4, 142)]]
first = []
second = []
for l in lst:
for tup in l:
first.append(tup[:-1])
second.append((tup[-1],))
print(first)
# [(1, 1), (1, 2), (0, 5), (0, 6), (2, 1), (1, 2), (10, 3), (2, 4)]
print(second)
# [(96,), (95,), (23,), (22,), (145,), (144,), (143,), (142,)]
或使用列表理解:
first = [tup[:-1] for l in lst for tup in l]
second = [(tup[-1],) for l in lst for tup in l]
然后将这些列表分别转换为2个子列表:
sublen = len(lst[0])
def split_lists(l, s):
return [l[i:i+s] for i in range(0, len(l), s)]
print(split_lists(first, sublen))
# [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3), (2, 4)]]
print(split_lists(second, sublen))
# [[(96,), (95,), (23,), (22,)], [(145,), (144,), (143,), (142,)]]
只需保持简单,循环嵌套列表,并获取所需内容:
lst = [[(1, 1, 96),
(1, 2, 95),
(0, 5, 23),
(0, 6, 22)],
[(2, 1, 145),
(1, 2, 144),
(10, 3, 143),
(2, 4, 142)]]
first = []
second = []
for l in lst:
for tup in l:
first.append(tup[:-1])
second.append((tup[-1],))
print(first)
# [(1, 1), (1, 2), (0, 5), (0, 6), (2, 1), (1, 2), (10, 3), (2, 4)]
print(second)
# [(96,), (95,), (23,), (22,), (145,), (144,), (143,), (142,)]
或使用列表理解:
first = [tup[:-1] for l in lst for tup in l]
second = [(tup[-1],) for l in lst for tup in l]
然后将这些列表分别转换为2个子列表:
sublen = len(lst[0])
def split_lists(l, s):
return [l[i:i+s] for i in range(0, len(l), s)]
print(split_lists(first, sublen))
# [[(1, 1), (1, 2), (0, 5), (0, 6)], [(2, 1), (1, 2), (10, 3), (2, 4)]]
print(split_lists(second, sublen))
# [[(96,), (95,), (23,), (22,)], [(145,), (144,), (143,), (142,)]]
尽管Amadan的答案很好,但我还是想编写一个通用函数来实现所需的结果。以下是最终代码:-
def fn(data, one_shot_columns, scalar_columns):
zipped=list(zip(*data))
one_shot_zipped=[zipped[i] for i in one_shot_columns]
one_shot=list(zip(*one_shot_zipped))
scalar_zipped=[zipped[i] for i in scalar_columns]
scalar=list(zip(*scalar_zipped))
return (one_shot,scalar)
用法如下:
hist_oneshot,hist_scalar = fn(hist,[0,1],[2])
尽管Amadan的答案很好,但我还是想编写一个通用函数来实现所需的结果。以下是最终代码:-
def fn(data, one_shot_columns, scalar_columns):
zipped=list(zip(*data))
one_shot_zipped=[zipped[i] for i in one_shot_columns]
one_shot=list(zip(*one_shot_zipped))
scalar_zipped=[zipped[i] for i in scalar_columns]
scalar=list(zip(*scalar_zipped))
return (one_shot,scalar)
用法如下:
hist_oneshot,hist_scalar = fn(hist,[0,1],[2])
(96)
(相当于96
,一个整数)或(96,)
(一元组)?(96,)相当于一元组。(96)
不等同于一元组(96)
只是括号中的一个整数,相当于(95+1)
(96,)
是一个1元组。参见:(96)+(1)
是97
<代码>(96,)+(1,)是(96,1)
。感谢您指出这一点。我已经更新了问题和我的评论。(96)
(相当于96
,一个整数)或(96,)
(一个元组)?(96,)相当于一个元组。(96)
不等同于一个元组(96)
只是括号中的一个整数,相当于(95+1)
(96,)
是一个1元组。参见:(96)+(1)
是97
<代码>(96,)+(1,)是(96,1)
。感谢您指出这一点。我已经更新了问题和我的评论。这很有效!此外,还有一种方法可以编写某种类型的泛型函数,该函数接受输出中的列值。比如:-def get_first_x_columns(arr,columns),这里列的值是(0,1)。不确定您的意思,您应该展示一个如何使用该函数的示例。基本上,如果我添加其他列,那么您答案中的以下命令将变成:-[[(c,d)表示a,b,c,*d in r]表示r in arr]。所以,我希望它是可配置的,这样每次添加一个新列时,就可以避免这种情况。在函数中,我可以为前两列传递[arr,(0,1)],为下两列传递[arr,(2,3)]。这非常有效!此外,还有一种方法可以编写某种类型的泛型函数,该函数接受输出中的列值。比如:-def get_first_x_columns(arr,columns),这里列的值是(0,1)。不确定您的意思,您应该展示一个如何使用该函数的示例。基本上,如果我添加其他列,那么您答案中的以下命令将变成:-[[(c,d)表示a,b,c,*d in r]表示r in arr]。所以,我希望它是可配置的,这样每次添加一个新列时,就可以避免这种情况。如中所示,在函数中,我可以为前2列传递[arr,(0,1)],为下2列传递[arr,(2,3)]。