Python 对数组中的特定字母进行计数

我被要求创建一个函数，在该函数中，它统计一个字符串中来自b-s的所有删除者。以下是我到目前为止的情况：

def strange_count(s):
    count = 0
    s = s.lower()
    s_count = [b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s]
    while count <= len(s):
        for i in range(len(s_count)):
            if s_count[i] == s[count]:
                count += 1
    return      

---

您可以使用字母按字典顺序排序的事实，而不是制作列表，这样您就可以将它们与

和

='b'和i这一行进行比较：

s_count = [b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s]

应该是这样的：

s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]

def strange_count(s):
  count   = 0
  s       = s.lower()
  s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]
  for letter in s:
    if letter in s_count:
      count += 1
  return count

print(strange_count("asfdsjfdlkwjrwoiureaoifhwabrejwer"))

input = "Hello world"
input = input.lower()
list_chars = ['b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s']
counts = {}
for c in list_chars:
    counts[c] = input.count(c)

print counts

然后你会发现你处于一个无限循环中，所以试着像这样重写：

s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]

def strange_count(s):
  count   = 0
  s       = s.lower()
  s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]
  for letter in s:
    if letter in s_count:
      count += 1
  return count

print(strange_count("asfdsjfdlkwjrwoiureaoifhwabrejwer"))

input = "Hello world"
input = input.lower()
list_chars = ['b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s']
counts = {}
for c in list_chars:
    counts[c] = input.count(c)

print counts

问题-变量与字母
上面是而不是声明了一个列表，其中包含与您正在谈论的字母范围相对应的元素

你真正想说的是
s\u count
应该是一个列表，其中元素的值应该对应于变量
b
，
c
，
d
，
e
，
f
，
g
，
h
，
i
，
j
，
k
，
l
，
m
，
n
，
o
，
p
，
q
，
r
和
s

您将从python解释器获得一条诊断（错误消息），因为这些变量都没有声明


解决方案
如果要创建字符列表，则需要将它们包装在
“
或
”
（如下所示）中

注意：鉴于上述内容的编写和维护都非常繁琐，因此初始化
s_count
的更合适方法是
s_count=list（“bcdefghijklmnopqrs”）
——这将产生所需的结果

该错误表示您正在使用未定义的变量-
b
、
c
，等等。您似乎打算使用这些字符，在这种情况下，应将它们括在引号中：

s_count = ['b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s']

话虽如此，值得记住的是，字符是顺序的，因此您可以只使用
=
运算符。此外，IMHO，如果您只使用
求和
和一个列表，那么为其定义一个函数似乎有点过火了：

mystring = 'some string'
strange_count = sum(1 for c in s if c >= 'b' and c <= 's')

mystring='some string'
奇怪的_count=sum（如果c>='b'和c，则s中的c为1）在您的情况下，您可以查看每个字符是否>='b”和<='s“，指定某些字符的另一种有效方法是使用一组您要检查的字符，并将s中的字符在集合中的次数相加：

def strange_count(s):
    s = s.lower()
    s_count = {"b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s"}
    return sum(ch in s_count for ch in s)

就您的错误而言，您需要实际使用双引号或单引号创建字符串：

["b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s"]

使用
b，
c
…
python将查找使用
a，b，c…
名称定义的变量，因此您将得到
namererror
，除非您实际上在某个地方定义了这些变量

如果您要使用如果您没有处理很长的字符串，或者您没有被要求考虑计算成本，您可以使用如下内容：

s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]

def strange_count(s):
  count   = 0
  s       = s.lower()
  s_count = ["b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s"]
  for letter in s:
    if letter in s_count:
      count += 1
  return count

print(strange_count("asfdsjfdlkwjrwoiureaoifhwabrejwer"))

input = "Hello world"
input = input.lower()
list_chars = ['b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s']
counts = {}
for c in list_chars:
    counts[c] = input.count(c)

print counts

你需要担心大写字母吗？对于像“é”这样带变音符号的外语字母呢？