计算 Python 字符串中字符出现的次数

Question

我想获取给定句子中每个字符的数量。我尝试了下面的代码，我得到了每个字符的计数，但它在输出中显示了重复的字符计数。如何删除重复的字符。

def countwords(x):
    x=x.lower()
    for i in x:
        print(i,'in',x.count(i)) 

x=str(input("Enter a paragraph "))
countwords(x)

我的输出是：

我的输出不应包含空格数和重复字符..该怎么办....!!!

Answer 1

检查此代码：

my_string = "count a character occurance"
my_list = list(my_string)
print (my_list)
get_unique_char = set(my_list)
print (get_unique_char)

for key in get_unique_char:
    print (key, my_string.count(key))

Answer 2

有几种不同的方法，最有暗示的是jonrsharpe的评论，但我建议一个简单的

set

。

设置方法以及其他一些方法如下：

# An approach using a set
def countwords_set(s):
    for c in set(s):
        if c == ' ': continue
        print(c, 'in', s.count(c))

# An approach using a standard dict
def countwords_dict(s):
    d = dict()
    for c in s:
        if c == ' ': continue               # Skip spaces
        d[c] = d.get(c,0) + 1               # Use the .get method in case the 
                                            #   key isn't set

    for c,x in d.items():                   # Display results
        print(c, 'in', x)


# An approach using a defaultdict (from the collections module)
def countwords_ddict(s):
    from collections import defaultdict     # Typically, imports go at the top

    d = defaultdict(int)

    for c in s:
        if c == ' ': continue
        d[c] += 1

    for c,x in d.items():
        print(c, 'in', x)


# An approach using a Counter (from the collections module)
def countwords_counter(s):
    from collections import Counter         # Typically, imports go at the top

    counter = Counter(s)

    # Counters can be accessed like dicts
    for c,x in counter.items():
        if c == ' ': continue
        print(c, 'in', x)


# User input and comparison
s = str(input("Enter a paragraph "))
s = s.lower()

countwords_set(s)
print("---")

countwords_dict(s)
print("---")

countwords_ddict(s)
print("---")

countwords_counter(s)
print("---")

每种方法的输出本质上是相同的，尽管字符的顺序可能有所不同，因为 Python 字典是无序的。

Answer 3

使用字典。

def countwords(x):
    d = dict()
    x=x.lower()
    for i in x:
        if i in d.keys():
            d[i] = d[i] +1;
        else:
             d[i] = 1;

     for i in d.keys():
          print i + " " + d[i]

Answer 4

或者将 numpy 与此单行代码一起使用

np.unique(np.array(list('count these characters')), return_counts=True)

Answer 5

#Print the no of occurances of each char in a string including spaces and punctuations

    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)

输出：{'I': 2, 'n': 3, 'd': 2, 'i': 2, 'a': 2, ' ': 9, 'w': 2, 'o': 2 、“t”：2、“h”：2、“e”：3、“O”：1、“D”：1、“r”：2、“l”：1、“c”：1、“ u': 1、'p': 1、'y': 1、'2': 1、'0': 1、'1': 2、'!': 1}