python返回数组最大值索引 - 返回list中最大值的index




Python:在元组列表中找到最小值,最大值 (4)

一个普遍的方法是这样的:

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Python 3中都可用的函数:

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
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例子:alist的输出结果是

min_x = 1
max_x = 7

min_y = 3
max_y = 5

有没有简单的方法可以做?


使用枚举和列表理解的另一种解决方案

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

for num, k in enumerate(['X', 'Y']):
    print 'max_%s' %k, max([i[num] for i in alist])
    print 'min_%s' %k, min([i[num] for i in alist])

map(max, zip(*alist))

这首先解压你的列表,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意长度的元组。


>>> 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)




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