blocks|key|1009456|text|目前，您的输出是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1009457|the+key+is+<generator+object+<genexpr>+at+0x7fbaca5443c0>
the+val+is+<generator+object+<genexpr>+at+0x7fbaca5443c0>|code-block|syntax|javascript|1009458|这是因为k+for+k,v+in+dict是一个生成器表达式。不要把它和集合理解混淆起来，那些花括号是f字串的.|offset|length|style|CODE|1009459|当然，k+for+k,v+in+dict是有问题的。当您遍历字典本身时，它会给出键。因此，对于第一次迭代，"this"回来了。您不能将"this"解压缩为两个变量。k,+v+=+"this"。|1009460|你可以用这个：|1009461|d+=+{"this":+"I",+"is":+"am",+"title":+"done"}
print(f'the+key+is+{"+".join(d.keys())}')
print(f'the+val+is+{"+".join(d.values())}')|1009462|产出：|1009463|the+key+is+this+is+title
the+val+is+I+am+done|1009464|此join之所以有效，是因为键和值是字典中的字符串。如果它们不是，您应该将它们转换为：|1009465|print(f'the+key+is+{"+".join(map(str,+d.values()))}')|1009466|对于第一个版本，您也可以使用print(f'the+key+is+{"+".join(d)}')作为默认情况下，字典将在迭代中给出键。|1009467|entityMap^0|0|0|4|H|0|3|H|1H|6|1V|6|2A|D|0|0|0|0|0|1|4|0|0|E|Y|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@$I|1A|J|1B|K|L]|$I|1C|J|1D|K|L]|$I|1E|J|1F|K|L]|$I|1G|J|1H|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1K|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1M|8|@$I|1N|J|1O|K|L]]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1P|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1Q|8|@$I|1R|J|1S|K|L]]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|13|$]]

Currently your output is:
<pre class="lang-none prettyprint-override"><code>the key is &lt;generator object &lt;genexpr&gt; at 0x7fbaca5443c0&gt;
the val is &lt;generator object &lt;genexpr&gt; at 0x7fbaca5443c0&gt;
</code></pre>
That's because <code>k for k,v in dict</code> is a generator expression. Don't confuse it with set comprehension, those curly braces are for f-string.
But of course that <code>k for k,v in dict</code> is problematic. When you iterate over a dictionary itself, it gives you keys. So for the first iteration <code>&quot;this&quot;</code> comes back. you can't unpack <code>&quot;this&quot;</code> into two variables. <code>k, v = &quot;this&quot;</code>.
You can use this:
<pre class="lang-py prettyprint-override"><code>d = {&quot;this&quot;: &quot;I&quot;, &quot;is&quot;: &quot;am&quot;, &quot;title&quot;: &quot;done&quot;}
print(f'the key is {&quot; &quot;.join(d.keys())}')
print(f'the val is {&quot; &quot;.join(d.values())}')
</code></pre>
output:
<pre class="lang-none prettyprint-override"><code>the key is this is title
the val is I am done
</code></pre>
This <code>join</code> works because keys and values are strings in your dictionary. If they are not, you should convert them like:
<pre class="lang-py prettyprint-override"><code>print(f'the key is {&quot; &quot;.join(map(str, d.values()))}')
</code></pre>
For the first one you could also use <code>print(f'the key is {&quot; &quot;.join(d)}')</code> as dictionaries will give keys in the iteration by default.

blocks|key|1006639|text|我会这样做的，如果你不介意在最后有一个空间，我不是专家。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1006640|mydict+=+{"this":+"I",+"is":+"am",+"title":+"done"}


def+get_str(dict):
++++str_a+=+""
++++str_b+=+""
++++for+k+in+dict:
++++++++str_a+%2B=+f"{k}+"
++++++++str_b+%2B=+f"{dict[k]}+"
++++print(str_a)
++++print(str_b)


get_str(mydict)|code-block|syntax|javascript|1006641|产出：|1006642|this+is+title+
I+am+done+|1006643|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would do this, if you don't mind having a space at the end I'm no expert tho:
<pre><code>mydict = {&quot;this&quot;: &quot;I&quot;, &quot;is&quot;: &quot;am&quot;, &quot;title&quot;: &quot;done&quot;}


def get_str(dict):
 str_a = &quot;&quot;
 str_b = &quot;&quot;
 for k in dict:
 str_a += f&quot;{k} &quot;
 str_b += f&quot;{dict[k]} &quot;
 print(str_a)
 print(str_b)


get_str(mydict)
</code></pre>
output:
<pre><code>this is title 
I am done 
</code></pre>

blocks|key|86550|text|密钥的print("The+keys+are",+'+'.join(k+for+k+in+dict))|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|86551|值的print("The+keys+are",+'+'.join(v+for+k,+v+in+dict.items()))|86552|entityMap^0|3|1C|0|2|1N|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

<code>print(&quot;The keys are&quot;, ' '.join(k for k in dict))</code> for the keys
<code>print(&quot;The keys are&quot;, ' '.join(v for k, v in dict.items()))</code> for the values

I read a lot of tutorials, but can not find how to iterate dict in string.format like this:
<pre class="lang-py prettyprint-override"><code>dict = {'this':'I','is':'am','title':'done'}
print(f'the key is {k for k,v in dict}')
print(f'the val is {v for k,v in dict}')
</code></pre>
which I want a result like this:
<pre class="lang-none prettyprint-override"><code>the key is this is title
the val is I am done
</code></pre>
so I can print variable length dict.
<pre><code>dict = {'this':'I','is':'am','title':'done'}
print(f'the key is {k for k,v in dict}')
print(f'the val is {v for k,v in dict}')
</code></pre>
Then I got error.

How to iterate dict in string.format

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我读了很多教程，但是找不到如何在string.format中像这样迭代dict：dict = {'this':'I','is':'am','title':'done'}print(f'the key is {k for k,v in dict}')print(f'the val is {v for k,v in dic...

问如何在string.format中迭代dict
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问如何在string.format中迭代dictEN