迭代对应于Python中列表的字典键值

问题:迭代对应于Python中列表的字典键值

使用Python 2.7。我有一本字典,其中以球队名称为关键,对每支球队得分并允许的奔跑次数作为值列表:

NL_East = {'Phillies': [645, 469], 'Braves': [599, 548], 'Mets': [653, 672]}

我希望能够将字典提供给函数并遍历每个团队(键)。

这是我正在使用的代码。现在,我只能逐队参加。我将如何遍历每个团队并为每个团队打印预期的win_percentage?

def Pythag(league):
    runs_scored = float(league['Phillies'][0])
    runs_allowed = float(league['Phillies'][1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print win_percentage

谢谢你的帮助。

Working in Python 2.7. I have a dictionary with team names as the keys and the amount of runs scored and allowed for each team as the value list:

NL_East = {'Phillies': [645, 469], 'Braves': [599, 548], 'Mets': [653, 672]}

I would like to be able to feed the dictionary into a function and iterate over each team (the keys).

Here’s the code I’m using. Right now, I can only go team by team. How would I iterate over each team and print the expected win_percentage for each team?

def Pythag(league):
    runs_scored = float(league['Phillies'][0])
    runs_allowed = float(league['Phillies'][1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print win_percentage

Thanks for any help.


回答 0

您有几种选择可以遍历字典。

如果迭代字典本身(for team in league),则将迭代字典的键。当使用for循环进行循环时,无论您是在dict(league)本身上循环还是在以下情况下,行为都是相同的league.keys()

for team in league.keys():
    runs_scored, runs_allowed = map(float, league[team])

您还可以通过迭代遍历键和值一次league.items()

for team, runs in league.items():
    runs_scored, runs_allowed = map(float, runs)

您甚至可以在迭代时执行元组拆包:

for team, (runs_scored, runs_allowed) in league.items():
    runs_scored = float(runs_scored)
    runs_allowed = float(runs_allowed)

You have several options for iterating over a dictionary.

If you iterate over the dictionary itself (for team in league), you will be iterating over the keys of the dictionary. When looping with a for loop, the behavior will be the same whether you loop over the dict (league) itself, or league.keys():

for team in league.keys():
    runs_scored, runs_allowed = map(float, league[team])

You can also iterate over both the keys and the values at once by iterating over league.items():

for team, runs in league.items():
    runs_scored, runs_allowed = map(float, runs)

You can even perform your tuple unpacking while iterating:

for team, (runs_scored, runs_allowed) in league.items():
    runs_scored = float(runs_scored)
    runs_allowed = float(runs_allowed)

回答 1

您也可以很容易地遍历字典:

for team, scores in NL_East.iteritems():
    runs_scored = float(scores[0])
    runs_allowed = float(scores[1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print '%s: %.1f%%' % (team, win_percentage)

You can very easily iterate over dictionaries, too:

for team, scores in NL_East.iteritems():
    runs_scored = float(scores[0])
    runs_allowed = float(scores[1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print '%s: %.1f%%' % (team, win_percentage)

回答 2

字典具有一个称为的内置函数iterkeys()

尝试:

for team in league.iterkeys():
    runs_scored = float(league[team][0])
    runs_allowed = float(league[team][1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print win_percentage

Dictionaries have a built in function called iterkeys().

Try:

for team in league.iterkeys():
    runs_scored = float(league[team][0])
    runs_allowed = float(league[team][1])
    win_percentage = round((runs_scored**2)/((runs_scored**2)+(runs_allowed**2))*1000)
    print win_percentage

回答 3

字典对象允许您迭代其项目。此外,通过模式匹配和__future__可以使事情稍微简化。

最后,您可以将逻辑从打印中分离出来,以使事情在以后的重构/调试中更加容易。

from __future__ import division

def Pythag(league):
    def win_percentages():
        for team, (runs_scored, runs_allowed) in league.iteritems():
            win_percentage = round((runs_scored**2) / ((runs_scored**2)+(runs_allowed**2))*1000)
            yield win_percentage

    for win_percentage in win_percentages():
        print win_percentage

Dictionary objects allow you to iterate over their items. Also, with pattern matching and the division from __future__ you can do simplify things a bit.

Finally, you can separate your logic from your printing to make things a bit easier to refactor/debug later.

from __future__ import division

def Pythag(league):
    def win_percentages():
        for team, (runs_scored, runs_allowed) in league.iteritems():
            win_percentage = round((runs_scored**2) / ((runs_scored**2)+(runs_allowed**2))*1000)
            yield win_percentage

    for win_percentage in win_percentages():
        print win_percentage

回答 4

列表理解可以缩短内容…

win_percentages = [m**2.0 / (m**2.0 + n**2.0) * 100 for m, n in [a[i] for i in NL_East]]

List comprehension can shorten things…

win_percentages = [m**2.0 / (m**2.0 + n**2.0) * 100 for m, n in [a[i] for i in NL_East]]