计算字符串中字符的出现次数

Question

我被要求计算用户输入的字符串中字母、空格和标点符号的出现次数，并且在打印输出中，字母应按照它们在文本中出现的顺序出现，但任何字母都不应出现两次，而且小写和大写应该算作一个。到目前为止我的代码看起来像这样。

S = str(input("Enter a string: "))
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

for char in S:
     if char in alpha:
          count = S.count(char)
     print(char,':',count)

输出

Enter a string: hello
h : 1
e : 1
l : 2
l : 2
o : 1

我希望它看起来像这样

Enter a string: hello
    h : 1
    e : 1
    l : 2
    o : 1

我在想是否将字符串和出现的字符的计数转换成像这样的嵌套列表

Enter a string: hello
[['h', 1], ['e', 1], ['l', 2], ['l', 2], ['o', 1]]

我可以删除相同的列表并只留下一个吗？

Answer 1

>>> s = 'hello'
>>> s = s.lower()
>>> print('\n'.join('{} : {}'.format(c, s.count(c)) for i, c in enumerate(s) if c in "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! " and c not in s[:i]))
h : 1
e : 1
l : 2
o : 1

Answer 2

这是使用

Counter

模块的

collections

的经典情况：

from collections import Counter

S ='hello'
S = S.lower()
d = Counter(S)
for item, ct in d.items():
    print '%s occured %d times' % (item, ct)

Answer 3

只需迭代

set

S = str(input("Enter a string: ")).lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

for char in set(S):
    if char in alpha:
        print(char,':',S.count(char))

Answer 4

你可以这样做：

S = 'hello'
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

seen=set()

for c in S:
    if c not in seen and c in alpha:
        print '{} : {}'.format(c, S.count(c))
        seen.add(c)

打印：

h : 1
e : 1
l : 2
o : 1

Answer 5

所以我想出了如何在不使用集合和计数器的情况下做到这一点。

S = (input("Enter a string: "))
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "
L = []
#counts the occurrence of characters in a string.
#then puts them in a list
for char in S:
     if char in alpha:
          count = S.count(char)
     L.append(char)
     L.append(count)
#turns list into a nested list
L2 = []
i = 0
while i <len(L):
        L2.append(L[i:i+2])
        i += 2
#checks that the lists dont appear more than once 
L3 = []
for i in L2:
     if i not in L3:
          L3.append(i)

# print out the letters and count of the letter
for i,k in L3:
     print(i,k)

可能很长，但它有效。想听听您的意见吗？

Answer 6

这应该是问题的最佳解决方案

#打印字符串中每个字符出现的次数，包括空格和标点符号

text = "India won the ODI worldcup in the year 2011 !"
counter = dict()
for word in text:
    for char in word:
        count = text.count(char)
        counter[char] = count
print(counter)

字典排除了冗余键，并为我们提供了每个字符出现的单个键。