Question

我正在学习一些列表/词典理解，我被卡住了!!!

我真的不明白以下......

我有这个程序：

def histogram_for(word):
    adict = dict()
    for c in word:
        adict[c] = adict.get(c, 0) + 1
    print adict
    return adict

def histogram_dc(word):
    adict = dict()
    adict = {c: (adict.get(c, 0) + 1) for c in word}
    print adict
    return adict

def histogram_lc(word):
    adict = dict()
    adict = dict([(c, (adict.get(c, 0) + 1)) for c in word])
    print adict
    return adict


word = "Football"
histogram_for(word.lower())
histogram_dc(word.lower())
histogram_lc(word.lower())

我得到了这些结果：

{'a': 1, 'b': 1, 'f': 1, 'l': 2, 'o': 2, 't': 1}
{'a': 1, 'b': 1, 'f': 1, 'l': 1, 'o': 1, 't': 1}
{'a': 1, 'b': 1, 'f': 1, 'l': 1, 'o': 1, 't': 1}

为什么唯一有效的是“for”方法？

Answer 1

很简单，因为虽然_dc和_lc adict中的处理是空的，但在_for中for每次都会更新for }循环。理解可以在其自身的adict = dict() adict = {c: (adict.get(c, 0) + 1) for c in word}循环中消失：

adict = dict()
# Handwaving for what Python actually does
$newdict = {}
for c in word:
    $newdict[c] = adict.get(c, 0) + 1
adict = $newdict

变为：

getPageWidth()

如果您需要跟踪一组密钥和出现次数，请使用collections.Counter（或for循环版本）。

Answer 2

正如Sean Vieira所说，课程collections.Counter及其方法most_common是满足您需求的最佳解决方案。

但是，如果你真的想保持列表/字典理解，我建议使用set和count：

def histogram_dc(word):
  adict = dict()
  adict = {c: word.count(c) for c in set(word)}
  print adict
  return adict

def histogram_lc(word):
  adict = dict()
  adict = dict([(c, word.count(c)) for c in set(word)])
  print adict
  return adict

python和list / dict理解

2 个答案: