blocks|key|2699099|text|不是的。使用迭代器时，您对位置一无所知-实际上，迭代器可以是无限的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2699100|除此之外，字典是not+ordered。因此，如果您需要它，例如在元素之间插入逗号，您应该获取这些项，对它们进行排序，然后在(key,+value)元组列表中迭代它们。当遍历这个列表时，你可以很容易地计算出迭代的次数，从而知道什么时候你有最后一个元素。|offset|length|style|BOLD|CODE|2699101|entityMap^0|0|8|B|1Q|C|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@$D|M|E|N|F|G]|$D|O|E|P|F|H]]|9|@]|A|$]]|$1|I|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|J|$]]

No. When using an iterator you do not know anything about the position - actually, the iterator could be infinite.

Besides that, a dictionary is not ordered. So if you need it e.g. to insert commas between the elements you should take the items, sort them and them iterate over the list of <code>(key, value)</code> tuples. And when iterating over this list you can easily count the number of iterations and thus know when you have the last element.

blocks|key|2415409|text|有一种丑陋的方法可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2415410|for+i,+(k,+v)+in+enumerate(your_dict.iteritems()):
++++if+i+==+len(your_dict)-1:
++++++++#+do+special+stuff+here|code-block|syntax|javascript|2415411|但是你真的应该考虑一下你是否需要这个。我几乎可以肯定还有另一种方法。|2415412|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

There is an ugly way to do this:

<pre><code>for i, (k, v) in enumerate(your_dict.iteritems()):
 if i == len(your_dict)-1:
 # do special stuff here
</code></pre>

But you should really consider if you need this. I am almost certain that there is another way.

blocks|key|2699159|text|it+=+spam_dict.iteritems()
try:
++++eggs1+=+it.next()
++++while+True:
++++++++eggs2+=+it.next()
++++++++do_something(eggs1)
++++++++eggs1+=+eggs2
except+StopIteration:
++++do_final(eggs1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2699160|又快又脏。它能解决你的问题吗？|unstyled|2699161|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>it = spam_dict.iteritems()
try:
 eggs1 = it.next()
 while True:
 eggs2 = it.next()
 do_something(eggs1)
 eggs1 = eggs2
except StopIteration:
 do_final(eggs1)
</code></pre>

Quick and quite dirty. Does it solve your issue?

blocks|key|797557|text|正如其他人所说的，字典没有定义的顺序，所以很难想象为什么需要它，但这里就是它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|797558|last+=+None
for+current+in+your_dict.iteritems():
++if+last+is+not+None:
++++#+process+last
++last+=+current

#+now+last+contains+the+last+thing+in+dict.iteritems()
if+last+is+not+None:+#+this+could+happen+if+the+dict+was+empty
++#+process+the+last+item|code-block|syntax|javascript|797559|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

as others have stated, dictionaries have no defined order, so it's hard to imagine why you would need this, but here it is

<pre><code>last = None
for current in your_dict.iteritems():
 if last is not None:
 # process last
 last = current

# now last contains the last thing in dict.iteritems()
if last is not None: # this could happen if the dict was empty
 # process the last item
</code></pre>

blocks|key|2415487|text|最有意义的方法是将循环包装在某个调用中，该调用包含一个钩子，以便在之后调用您的迭代后功能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2415488|这可以作为上下文管理器实现，并通过'with‘语句调用，或者，对于旧版本的Python，您可以使用旧的'try:’...‘最后：’构造。它也可以包装在一个类中，在这个类中，字典迭代是自调度的(一个“私有”方法)，附录代码紧跟在公共方法中。(理解公共和私有之间的差异是一个意图和文档问题，而不是由Python强制执行的)。|2415489|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

The approach that makes the most sense is to wrap the loop in some call which contains a hook to call your post-iteration functionality afterwards.

This could be implemented as context manager and called through a 'with' statement or, for older versions of Python, you could use the old 'try:' ... 'finally:' construct. It could also be wrapped in a class where the dictionary iteration is self dispatched (a "private" method) and the appendix code follows that in the public method. (Understanding that the distension between public vs private is a matter of intention and documentation, not enforced by Python).

blocks|key|2415525|text|我最近遇到了这个问题，我认为这是最优雅的解决方案，因为它允许您编写for+i,value,isLast+in+lastEnumerate(...):|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2415526|def+lastEnumerate(iterator):
++++x+=+list(iterator)
++++for+i,value+in+enumerate(x):
++++++++yield+i,value,i==len(x)-1|code-block|syntax|javascript|2415527|例如：|2415528|for+i,value,isLast+in+lastEnumerate(range(5)):
++++print(value)
++++if+not+isLast:
++++++++print(',')|2415529|entityMap^0|X|15|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I recently had this issue, I thought this was the most elegant solution because it allowed you to write <code>for i,value,isLast in lastEnumerate(...):</code>:

<pre><code>def lastEnumerate(iterator):
 x = list(iterator)
 for i,value in enumerate(x):
 yield i,value,i==len(x)-1
</code></pre>

For example:

<pre><code>for i,value,isLast in lastEnumerate(range(5)):
 print(value)
 if not isLast:
 print(',')
</code></pre>

blocks|key|2699316|text|这是this+broader+question的一个特例。我的建议是创建一个对最后一项返回-1的enumerate-like+generator：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2699317|def+annotate(gen):
++++prev_i,+prev_val+=+0,+gen.next()
++++for+i,+val+in+enumerate(gen,+start=1):
++++++++yield+prev_i,+prev_val
++++++++prev_i,+prev_val+=+i,+val
++++yield+'-1',+prev_val|code-block|syntax|javascript|2699318|如果您希望它处理序列和生成器，则添加gen+=+iter(gen)。|style|CODE|2699319|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/5983265/577088|1|https://stackoverflow.com/questions/5983265/pythonic-way-of-determining-if-the-current-element-is-the-first-or-last-element-o/5983613#5983613^0|2|L|0|1C|O|1|0|0|I|F|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|14|8|@$A|15|B|16|K|L]]|9|@]|C|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

This is a special case of <a href="https://stackoverflow.com/q/5983265/577088">this broader question</a>. My suggestion was to create an <a href="https://stackoverflow.com/questions/5983265/pythonic-way-of-determining-if-the-current-element-is-the-first-or-last-element-o/5983613#5983613">enumerate-like generator</a> that returns -1 on the last item:

<pre><code>def annotate(gen):
 prev_i, prev_val = 0, gen.next()
 for i, val in enumerate(gen, start=1):
 yield prev_i, prev_val
 prev_i, prev_val = i, val
 yield '-1', prev_val
</code></pre>

Add <code>gen = iter(gen)</code> if you want it to handle sequences as well as generators.

blocks|key|2415607|text|无论如何，for循环中的最后一项会在for循环之后挂起：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2415608|for+current_item+in+my_dict:
++++do_something(current_item)

try:
++++do_last(current_item)
except+NameError:
++++print+"my_dict+was+empty"|code-block|syntax|javascript|2415609|即使在for循环之前使用了名称"current_item“，尝试在空字典上循环似乎也会删除current_item，因此NameError|2415610|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The last item in a for loop hangs around after the for loop anyway:

<pre><code>for current_item in my_dict:
 do_something(current_item)

try:
 do_last(current_item)
except NameError:
 print "my_dict was empty"
</code></pre>

Even if the name "current_item" is in use before the for loop, attempting to loop over an empty dict seems to have the effect of deleting current_item, hence the NameError

blocks|key|4491055|text|您在上面的注释中指出，您需要它来构造SQL+SELECT语句的WHERE子句。也许这会有所帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4491056|def+make_filter(colname,+value):
++++if+isinstance(value,+str):
++++++++if+'%25'+in+value:
++++++++++++return+"%25s+LIKE+'%25s'"+%25+(colname,+value)
++++++++else:
++++++++++++return+"%25s+=+'%25s'"+%25+(colname,+value)
++++return+"%25s+=+%25s"+%25+(colname,+value)

filters+=+{'USER_ID':'123456',+'CHECK_NUM':23459,+'CHECK_STATUS':'C%25'}
whereclause+=+'WHERE+'%2B'\nAND+'.join(make_filter(*x)+for+x+in+filters.iteritems())
print+whereclause|code-block|syntax|javascript|4491057|哪种打印|4491058|WHERE+CHECK_NUM+=+23459
AND+CHECK_STATUS+LIKE+'C%25'
AND+USER_ID+=+'123456'|4491059|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|$]]

You stated in an above comment that you need this to construct the WHERE clause of an SQL SELECT statement. Perhaps this will help:

<pre><code>def make_filter(colname, value):
 if isinstance(value, str):
 if '%' in value:
 return "%s LIKE '%s'" % (colname, value)
 else:
 return "%s = '%s'" % (colname, value)
 return "%s = %s" % (colname, value)

filters = {'USER_ID':'123456', 'CHECK_NUM':23459, 'CHECK_STATUS':'C%'}
whereclause = 'WHERE '+'\nAND '.join(make_filter(*x) for x in filters.iteritems())
print whereclause
</code></pre>

which prints

<pre><code>WHERE CHECK_NUM = 23459
AND CHECK_STATUS LIKE 'C%'
AND USER_ID = '123456'
</code></pre>

blocks|key|797762|text|我知道这一点很晚，但我是如何解决这个问题的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|797763|dictItemCount+=+len(dict)
dictPosition+=+1
for+key,value+in+dict
++++if(dictPosition+=+dictItemCount):
++++++++print+'last+item+in+dictionary'
++++dictPosition+%2B=+1|code-block|syntax|javascript|797764|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I know this late, but here's how I've solved this issue:

<pre><code>dictItemCount = len(dict)
dictPosition = 1
for key,value in dict
 if(dictPosition = dictItemCount):
 print 'last item in dictionary'
 dictPosition += 1
</code></pre>

blocks|key|2699455|text|另一种方法是枚举您的字典，并将当前迭代与最终迭代进行比较。在我看来，它更容易查看和理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2699456|for+n,+(key,+value)+in+enumerate(yourDict.items()):
++++if+yourDict[n]+==+yourDict[-1]:
++++++++print('Found+the+last+iteration!:',+n)|code-block|syntax|javascript|2699457|或者你可以在迭代结束后做一些事情：|2699458|for+key,+value+in+yourDict.items():
++++pass
else:
++++print('Finished+iterating+over+`yourDict`')|2699459|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|$]]

Another approach is to enumerate your dict and compare the current iteration against the final one. Its easier to look at and understand in my opinion:

<pre><code>for n, (key, value) in enumerate(yourDict.items()):
 if yourDict[n] == yourDict[-1]:
 print('Found the last iteration!:', n)
</code></pre>

OR you could just do something once the iteration is finished:

<pre><code>for key, value in yourDict.items():
 pass
else:
 print('Finished iterating over `yourDict`')
</code></pre>

Is there a simple way to detect the last iteration while iterating over a dictionary using <code>iteritems()</code>?

Detect last iteration over dictionary.iteritems() in python

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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在使用iteritems()遍历字典时，有没有一种简单的方法来检测最后一次迭代

问检测python中dictionary.iteritems()的最后一次迭代
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问检测python中dictionary.iteritems()的最后一次迭代EN