<p>我知道那本字典没有订。我知道</p>
<p>然而,这是一个计算词频的代码</p>
<pre><code>def wordCount(kalimat):
counter = {}
for kata in kalimat.split(" "):
if kata in counter:
counter[kata] += 1
else:
counter[kata] = 1
for I in sorted(counter):
if counter[I] == 1:
print("{:<10} appears 1 time.".format(I,counter[I]))
else:
print("{:<10} appears {:<3} times.".format(I,counter[I]))
</code></pre>
<p>我用下面的字符串调用了wordCount</p>
<blockquote>
<p>A word may appear once but twice may appear twice since this one will not appear again with this animal</p>
</blockquote>
<p>这就是结果</p>
<p>运行#1</p>
<pre><code>again appears 1 time.
not appears 1 time.
one appears 1 time.
may appears 2 times.
word appears 1 time.
appear appears 3 times.
since appears 1 time.
twice appears 2 times.
but appears 1 time.
with appears 1 time.
will appears 1 time.
A appears 1 time.
animal appears 1 time.
this appears 2 times.
once appears 1 time.
</code></pre>
<p>运行#2</p>
<pre><code>once appears 1 time.
word appears 1 time.
will appears 1 time.
animal appears 1 time.
appear appears 3 times.
again appears 1 time.
A appears 1 time.
not appears 1 time.
one appears 1 time.
but appears 1 time.
twice appears 2 times.
may appears 2 times.
with appears 1 time.
since appears 1 time.
this appears 2 times.
</code></pre>
<p>我明白这不是命令,但即使他们没有命令,为什么命令是不同的?我的想象力是它不按字母顺序排列的原因,因为顺序是基于它们注册的时间(即队列)</p>
<p>我无法想象当我想显示它时他们会调用random.shuffle()</p>