Python day-Dictionary，使用嵌套循环的字符频率

字典-{}

-->字典用于以键：值对的形式存储数据值。
-->字典是一个有序的、可更改的、不允许重复的集合。
-->在字典中，每个元素都可以通过它们的键来访问，而不是通过索引。
-->如果字典不包含该键，则输出将为“KeyError”。

示例：

thisdict = {
  "brand": "Ford",
  "model": "Mustang",
  "year": 1964
}

student = {"name":"raja", "class":5}

print(thisdict)
print(student)

输出：

{'brand': 'Ford', 'model': 'Mustang', 'year': 1964}
{'name': 'raja', 'class': 5}

练习：使用嵌套循环查找字符串中的字符
1.查找字符串中每个字母的频率

s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name print j>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">g 1
u 2
r 2
p 1
a 3
s 1
n 2

*2。字母仅出现一次*

s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and print j>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">g 1
p 1
s 1

3。最常见的字母

s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and print j>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">u 2
r 2
a 3
n 2

4。第一个不重复的字母

s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and print break j>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">g 1

5。第一个重复的字母

s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and print break j>



<p><strong>6。最后一个不重复的字母</strong><br>
</p>

<pre class="brush:php;toolbar:false">last = ' '
last_count = 0 
s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and last="key" last_count="count" j print>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">s 1

7。最后重复的字母

last = ' '
last_count = 0 
s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" count="1" i="j+1" if while name and last="key" last_count="count" j print>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">n 2

8。最常见的字母

s = 'guruprasanna'
name = list(s)
j = 0 
last = ' '
last_count = 0 

while j<len key="name[j]" count="1" i="j+1" if while name and>last_count:
            last = key
            last_count = count
    j+=1

print(last, last_count)
</len>

9。元音出现频率 (a,e,i,o,u)

vowels = ['a','e','i','o','u']
last = ' '
last_count = 0 
s = 'guruprasanna'
name = list(s)
j = 0 
while j<len key="name[j]" if in vowels: count="1" i="j+1" while name print j>



<p>输出：<br>
</p>

<pre class="brush:php;toolbar:false">u 2
a 3

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