循环遍历字典以按值顺序显示键和值

Question

我试图用for循环遍历字典，以按其中一个值元素(击球顺序号)的顺序显示键和值。

我可以以预期的格式打印键和值，但我不知道如何使这些行按正确的顺序排列。

我需要在for循环和中使用逻辑完成此任务，而不使用lambda或函数。这是我目前所做的:

print ('Rays starters' + "n")
rays_starters = {
    'DeJesus' : ['DH', 6, 299],
    'Loney' : ['1B', 4, 222],
    'Rivera' : ['C', 9, 194],
    'Forsythe' : ['2B', 5, 304],
    'Souza Jr' : ['RF', 2, 229],
    'Longoria' : ['3B', 3, 282],
    'Cabrera' : ['SS', 7, 214],
    'Kiermaier' : ['CF', 1, 240],
    'Guyer' : ['LF', 8, 274] }
for player in rays_starters:
    print (player + str(rays_starters[player]))
print ('n' + 'Today's lineup' + 'n')
for player in rays_starters:
    batting_order = rays_starters.get(player)
    print('Batting ' + str(batting_order[1]) + ' : ' + str(batting_order[0]) + ' ' + player + ' ,current avg: ' + str(batting_order[2]))   

输出应该像这样:

Rays starters
DeJesus ['DH', 6, 299]
Loney ['1B', 4, 222]
Rivera ['C', 9, 194]
Forsythe ['2B', 5, 304]
Souza Jr ['RF', 2, 229]
Longoria ['3B', 3, 282]
Cabrera ['SS', 7, 214]
Kiermaier ['CF', 1, 240]
Guyer ['LF', 8, 274]
Today's lineup
Batting 1 : CF Kiermaier ,current avg: 240
Batting 2 : RF Souza Jr ,current avg: 229
Batting 3 : 3B Longoria ,current avg: 282
Batting 4 : 1B Loney ,current avg: 222
Batting 5 : 2B Forsythe ,current avg: 304
Batting 6 : DH DeJesus ,current avg: 299
Batting 7 : SS Cabrera ,current avg: 214
Batting 8 : LF Guyer ,current avg: 274
Batting 9 : C Rivera ,current avg: 194

我的输出实际上看起来完全像这样，除了击球顺序出了问题。请帮助我走上正确的轨道，记住我正在努力学习，所以任何有用的批评是欢迎的!

Answer 1

这是一个有效的方法，因为我们知道将有9个击球手。

lineup = [None] * 9
for player, stats in rays_starters.items():
    lineup[stats[1]-1] = player, stats
print ('nToday's lineupn')
for player, batting_order in lineup:
    print('Batting ' + str(batting_order[1]) + ' : ' + str(batting_order[0]) + ' ' + player + ' ,current avg: ' + str(batting_order[2]))

我们所做的只是初始化一个包含9个元素的数组，并使用击球顺序将球员和统计数据作为元组映射到正确的数组索引。然后循环遍历球员和统计元组数组，并打印所需的格式化输出。这是O(n)

这个概念基本上来源于基数排序，或者更具体地说，是计数排序的一个非常简单的例子，其中所有的频率都是1，而"关键函数"只是从击球顺序中减去1来获得数组索引。

正如@PadraicCunningham在评论中指出的那样，理论上可以通过使用len函数来用于任何数量的电池。

Answer 2

字典没有顺序，因此无法对它们进行排序。但是，您可以以排序的方式迭代其值。为此，您可以使用sorted()和一个键函数，该函数指定在传递(key, value)元组时如何获取值:

for player, batting in sorted(rays_starters.items(), key=lambda x: x[1][1]):
    print('Batting {1} : {0} {player}, current avg: {2}'.format(*batting, player=player))

对于rays_starters字典，这将产生以下结果:

Batting 1 : CF Kiermaier, current avg: 240
Batting 2 : RF Souza Jr, current avg: 229
Batting 3 : 3B Longoria, current avg: 282
Batting 4 : 1B Loney, current avg: 222
Batting 5 : 2B Forsythe, current avg: 304
Batting 6 : DH DeJesus, current avg: 299
Batting 7 : SS Cabrera, current avg: 214
Batting 8 : LF Guyer, current avg: 274
Batting 9 : C Rivera, current avg: 194

如果不能指定这样的键函数，则必须自己实现排序。为此，您可以首先将字典转换为列表，然后对其进行排序。为了不需要键函数，您应该构造该列表，以便您要排序的值是列表中的第一个:

data = []
for player, batting in rays_starters.items():
    data.append((batting[1], player, batting[0], batting[2]))
# now sort the list
data.sort()
# and iterate and print
for player in data:
    print('Batting {0} : {1} {2}, current avg: {3}'.format(*player))

您也可以使用列表推导式创建data列表:

data = [(b[1], p, b[0], b[2]) for p, b in rays_starters.items()]

Answer 3

一种非常简单但效率很低的方法是在数组上迭代len(rays_starters)次数，并且每次只打印与顺序匹配的结果。

for i in range (1, len(rays_starters)+1):
    for player in rays_starters:
        if rays_starters[player][1] == i:
            print('Batting ' + str(rays_starters[player][1]) + ' : ' + 
            rays_starters[player][1] + ' ' + player + ' ,current avg: ' + 
            rays_starters[player][1])

Answer 4

如果你不能使用min, lambdas, sorted和其他函数调用等，手动查找具有最低击球数的球员:

out = []
cp = rays_starters.copy()
# keep going while the dict is not empty
while cp:
    mn = float("inf")
    it = None
    # iterate over the item to find the min each time
    # from remaining items
    for k, v in cp.items():
        if v[1] < mn:
            mn = v[1]
            it = (k, v)
    # append current it k/v pair which has the lowest 
    # batting number         
    out.append(it)
    # remove the key so we can get the next lowest
    del cp[it[0]]
for k,v in out:
    print("{} {}".format(k,v))

输出:

Kiermaier ['CF', 1, 240]
Souza Jr ['RF', 2, 229]
Longoria ['3B', 3, 282]
Loney ['1B', 4, 222]
Forsythe ['2B', 5, 304]
DeJesus ['DH', 6, 299]
Cabrera ['SS', 7, 214]
Guyer ['LF', 8, 274]
Rivera ['C', 9, 194]

或不复制:

out = []
seen = set()
# for every player in the dict
for _ in rays_starters):
    mn = float("inf")
    it = None
    # again get min each time based on the batting number
    for k, v in rays_starters.items():
        # this time we make sure we have not already used
        # the player 
        if v[1] < mn and k not in seen:
            mn = v[1]
            it = (k, v)
    out.append(it)
    # add the name of the player that matches our current min
    seen.add(it[0])
for k,v in out:
    print("{} {}".format(k,v))

如果你真的可以排序或者使用sorted使用batting number作为键，只对项目进行排序:

temp = {v[1]:[k]+v for k, v in rays_starters.items()}
for k ,v in sorted(temp.items()):
    print("{} {}".format(v[0], v[1:]))
 Kiermaier ['CF', 1, 240]
Souza Jr ['RF', 2, 229]
Longoria ['3B', 3, 282]
Loney ['1B', 4, 222]
Forsythe ['2B', 5, 304]
DeJesus ['DH', 6, 299]
Cabrera ['SS', 7, 214]
Guyer ['LF', 8, 274]
Rivera ['C', 9, 194]

如果击球总是从1开始:

temp = {v[1]:[k]+v for k, v in rays_starters.items()}
for k  in range(1,len(rays_starters)+1):
    v = temp[k]
    print("{} {}".format(v[0], list(v[1:])))

可以解包打印:

temp = {v[1]:[k]+v for k, v in rays_starters.items()}
for k in range(1,len(rays_starters)+1):
    name, nm, btn, avg = temp[k]
    print("Batting: {} {} {}, current avg: {}".format(btn, name, nm, avg))

输出:

Batting: 1 Kiermaier CF, current avg: 240
Batting: 2 Souza Jr RF, current avg: 229
Batting: 3 Longoria 3B, current avg: 282
Batting: 4 Loney 1B, current avg: 222
Batting: 5 Forsythe 2B, current avg: 304
Batting: 6 DeJesus DH, current avg: 299
Batting: 7 Cabrera SS, current avg: 214
Batting: 8 Guyer LF, current avg: 274
Batting: 9 Rivera C, current avg: 194

Answer 5

似乎一个更好的方法来处理这将是一个(排序)列表的namedtuples

from collections import namedtuple
Player = namedtuple("Player", ['name', 'position', 'order', 'avg'])
players = [Player(*info) for info in [("DeJesus", "DH", 6, 299),
                                      ('Loney', '1B', 4, 222),
                                      ...]]
players.sort(key=lambda p: p.order)
for player in players:
    print("Batting {p.order} : {p.position} {p.name}, current avg {p.avg}".format(p=p))

Answer 6

我迟到了，但我还是想添加这个hack:

>>> for x in {v[1]:'Batting {2} : {1} {0} ,current avg: {3}'.format(k,*v)
              for k,v in rays_starters.items()}.values():
        print(x)
Batting 1 : CF Kiermaier ,current avg: 240
Batting 2 : RF Souza Jr ,current avg: 229
Batting 3 : 3B Longoria ,current avg: 282
Batting 4 : 1B Loney ,current avg: 222
Batting 5 : 2B Forsythe ,current avg: 304
Batting 6 : DH DeJesus ,current avg: 299
Batting 7 : SS Cabrera ,current avg: 214
Batting 8 : LF Guyer ,current avg: 274
Batting 9 : C Rivera ,current avg: 194

但是，不要这样做，它依赖于字典项的实际顺序，而不是保证的顺序。我把它贴出来只是为了娱乐。

另一方面，它仍然工作，如果你删除一些播放器，不像接受的答案:-P