Question

alist = [(1,3),(2,5),(2,4),(7,5)]

我需要获取元组中每个位置的最小最大值。

Fox示例： alist的预期输出是

min_x = 1
max_x = 7

min_y = 3
max_y = 5

有简单的方法吗？

Answer 1

map(max, zip(*alist))

首先解压缩您的列表，然后找到每个元组位置的最大值

>>> alist = [(1,3),(2,5),(2,4),(7,5)]
>>> zip(*alist)
[(1, 2, 2, 7), (3, 5, 4, 5)]
>>> map(max, zip(*alist))
[7, 5]
>>> map(min, zip(*alist))
[1, 3]

这也适用于列表中任何长度的元组。

Answer 2

>>> from operator import itemgetter
>>> alist = [(1,3),(2,5),(2,4),(7,5)]
>>> min(alist)[0], max(alist)[0]
(1, 7)
>>> min(alist, key=itemgetter(1))[1], max(alist, key=itemgetter(1))[1]
(3, 5)

Answer 3

一种通用的方法是这样的：

alist = [(1,6),(2,5),(2,4),(7,5)]

temp = map(sorted, zip(*alist))
min_x, max_x, min_y, max_y = temp[0][0], temp[0][-1], temp[1][0], temp[1][-1]

对于Python 3，您需要将创建temp的行更改为：

temp = tuple(map(sorted, zip(*alist)))

这个想法可以被抽象成一个在Python 2和3中都有效的函数：

from __future__ import print_function
try:
    from functools import reduce  # moved into functools in release 2.6
except ImportError:
    pass

# readable version
def minmaxes(seq):
    pairs = tuple()
    for s in map(sorted, zip(*seq)):
        pairs += (s[0], s[-1])
    return pairs

# functional version
def minmaxes(seq):
    return reduce(tuple.__add__, ((s[0], s[-1]) for s in map(sorted, zip(*seq))))

alist = [(1,6), (2,5), (2,4), (7,5)]
min_x, max_x, min_y, max_y = minmaxes(alist)
print(' '.join(['{},{}']*2).format(*minmaxes(alist)))  # 1,7 4,6

triplets = [(1,6,6), (2,5,3), (2,4,9), (7,5,6)]
min_x, max_x, min_y, max_y, min_z, max_z = minmaxes(triplets)
print(' '.join(['{},{}']*3).format(*minmaxes(triplets)))  # 1,7 4,6 3,9

Answer 4

至少使用Python 2.7，“zip”不是必需的，因此这简化为map(max, *data)（其中data是元组或相同长度列表的迭代器）。< / p>

Answer 5

使用枚举和列表理解的另一种解决方案

$ zipinfo IE11.zip 
Archive:  IE11.zip   5314224734 bytes   1 file
warning [IE11.zip]:  1019257298 extra bytes at beginning or within zipfile
  (attempting to process anyway)
error [IE11.zip]:  start of central directory not found;
  zipfile corrupt.
  (please check that you have transferred or created the zipfile in the
  appropriate BINARY mode and that you have compiled UnZip properly)

Answer 6

对于python 3：

alist = [(1,3),(2,5),(2,4),(7,5)]    
[x_range, y_range] = list(zip(map(min, *test_list), map(max, *alist)))

print(x_range, y_range) #prints: (1, 7) (3, 5)

由于zip / map返回迭代器<object at 0x00>，因此您需要使用list()

Python：在元组列表中查找最小值，最大值

6 个答案: