python - 计算列表中的元素会产生意外结果

Question

这个问题已经有答案了:

Loop "Forgets" to Remove Some Items [duplicate] (10 个回答)

已关闭 6 年前。

我正在尝试计算句子中每个字符的出现次数。我使用了下面的代码:

printed = False
sentence = "the quick brown fox jumps over the lazy dog"
chars = list(sentence)
count = 0
for char in chars:
    if char == ' ':
        chars.remove(char)
    if printed == False:    
        count = chars.count(char)
        print "char count: ", char, count
    else:
        printed = False  

问题是，除了第一个单词之外，每个单词的第一个字母都没有打印，并且计数不正确(每次有新单词开始，计数就减 1，从 7 开始):

['t', 'h', 'e', ' ', 'q', 'u', 'i', 'c', 'k', ' ', 'b', 'r', 'o', 'w', 'n', ' ', 'f', 'o', 'x', ' ', 'j', 'u', 'm', 'p', 's', ' ', 'o', 'v', 'e', 'r', ' ', 't', 'h', 'e', ' ', 'l', 'a', 'z', 'y', ' ', 'd', 'o', 'g']
char count:  t 2
char count:  h 2
char count:  e 3
char count:    7
char count:  u 2
char count:  i 1
char count:  c 1
char count:  k 1
char count:    6
char count:  r 2
char count:  o 4
char count:  w 1
char count:  n 1
char count:    5
char count:  o 4
char count:  x 1
char count:    4
char count:  u 2
char count:  m 1
char count:  p 1
char count:  s 1
char count:    3
char count:  v 1
char count:  e 3
char count:  r 2
char count:    2
char count:  h 2
char count:  e 3
char count:    1
char count:  a 1
char count:  z 1
char count:  y 1
char count:    0
char count:  o 4
char count:  g 1

当我创建 2 个 for 循环(而不是 1 个)时，效果会更好:

sentence = "the quick brown fox jumps over the lazy dog"
chars = list(sentence)
count = 0
for char in chars:
    if char == ' ':
        chars.remove(char)
print chars
printed = False

for char in chars:
    if printed == False:
        count = chars.count(char)
        print "char count: ", char, count
        printed = True
    else:
        printed = False

这是输出:

['t', 'h', 'e', 'q', 'u', 'i', 'c', 'k', 'b', 'r', 'o', 'w', 'n', 'f', 'o', 'x', 'j', 'u', 'm', 'p', 's', 'o', 'v', 'e', 'r', 't', 'h', 'e', 'l', 'a', 'z', 'y', 'd', 'o', 'g']
char count:  t 2
char count:  e 3
char count:  u 2
char count:  c 1
char count:  b 1
char count:  o 4
char count:  n 1
char count:  o 4
char count:  j 1
char count:  m 1
char count:  s 1
char count:  v 1
char count:  r 2
char count:  h 2
char count:  l 1
char count:  z 1
char count:  d 1
char count:  g 1

唯一的问题是，“o”字符在输出中出现了两次...这是为什么？另外，为什么 1 循环不起作用？

Answer 1

迭代列表的副本并使用elif，否则在删除后继续，您不想计数空白处。

printed = False
sentence = "the quick brown fox jumps over the lazy dog"
chars = list(sentence)

for char in chars[:]:
    if char == ' ':
        chars.remove(char)
        continue
    if not printed:
        count = chars.count(char)
        print "char count: ", char, count
        printed = True
    else:
        printed = False

您也可以在按空格分割后仅str.join:

for char in "".join(sentence.split()):
    if not printed:
        count = chars.count(char)
        print "char count: ", char, count
        printed = True
    else:
        printed = False

但是，即使复制列表，您自己的解决方案实际上也无法正常工作，您在输出中丢失了字母:

char count:  t 2
char count:  e 3
char count:  u 2
char count:  c 1
char count:  b 1
char count:  o 4
char count:  n 1
char count:  o 4
char count:  j 1
char count:  m 1
char count:  s 1
char count:  v 1
char count:  r 2
char count:  h 2
char count:  l 1
char count:  z 1
char count:  d 1
char count:  g 1

字符串中有 26 个唯一字母，但您输出了 ~17 个。

您需要的是跟踪看到的字母并只打印一次计数，您的代码不会记录已打印的字符，它只是随机设置一个标志:

sentence = "the quick brown fox jumps over the lazy dog"
chars = list(sentence)

printed = set()
for char in "".join(sentence.split()):
    if char not in printed:
        count = chars.count(char)
        print "char count: ", char, count
        printed.add(char)

或者，如果第一次看到的顺序并不重要，那么只需在字符串上调用 set 即可:

for char in set("".join(sentence.split())):
    count = chars.count(char)
    print "char count: ", char, count

或者，如果您有大量数据，那么使用 Counter dict 会更好。 :

from collections import Counter
for char, count in Counter("".join(sentence.split())).items():
    print(char, count)