blocks|key|3042694|text|rows+=+list(myreader)
totalrows+=+len(rows)
for+i,+row+in+enumerate(rows):
++++print("Row+%25d/%25d"+%25+(i%2B1,+totalrows))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3042695|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>rows = list(myreader)
totalrows = len(rows)
for i, row in enumerate(rows):
 print("Row %d/%d" % (i+1, totalrows))
</code></pre>

blocks|key|306928|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306929|我找不到如何复制迭代器。|blockquote|306930|306931|306932|Closest是itertools.tee，但就像@J.F.Sebastian建议的那样，简单地对其进行list是最好的，正如itertools.tee的文档所解释的那样：|offset|length|style|CODE|306933|306934|此itertool可能需要大量的辅助存储(取决于需要存储多少临时数据)。通常，如果一个迭代器在另一个迭代器启动之前使用了大部分或所有数据，则使用list()而不是tee()会更快。|306935|306936|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/py3k/library/itertools.html?highlight=itertools.tee#itertools.tee^0|0|0|0|0|1G|4|8|D|0|0|0|20|6|29|5|0|0^^$0|@$1|2|3|-4|4|5|6|X|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|Y|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|10|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|11|7|@$H|12|I|13|J|K]]|8|@$H|14|I|15|1|16]]|9|$]]|$1|L|3|-4|4|5|6|17|7|@]|8|@]|9|$]]|$1|M|3|N|4|C|6|18|7|@$H|19|I|1A|J|K]|$H|1B|I|1C|J|K]]|8|@]|9|$]]|$1|O|3|-4|4|5|6|1D|7|@]|8|@]|9|$]]|$1|P|3|-4|4|5|6|1E|7|@]|8|@]|9|$]]]|Q|$R|$4|S|T|U|9|$V|W]]]]

<blockquote>
 I cannot find how to copy an
 iterator.
</blockquote>

Closest is <a href="http://docs.python.org/py3k/library/itertools.html?highlight=itertools.tee#itertools.tee" rel="nofollow noreferrer">itertools.tee</a>, but simply making a <code>list</code> of it, as @J.F.Sebastian suggests, is best here, as itertools.tee's docs explain:

<blockquote>
 This itertool may require significant
 auxiliary storage (depending on how
 much temporary data needs to be
 stored). In general, if one iterator
 uses most or all of the data before
 another iterator starts, it is faster
 to use <code>list()</code> instead of <code>tee()</code>.
</blockquote>

blocks|key|3256691|text|您只需打开该文件一次：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3256692|import+csv

f+=+open('myFile.csv',+'rb')

countrdr+=+csv.DictReader(f)
totalrows+=+0
for+row+in+countrdr:
++totalrows+%2B=+1

f.seek(0)++#+You+may+not+have+to+do+this,+I+didn't+check+to+see+if+DictReader+did

myreader+=+csv.DictReader(f)
for+row+in+myreader:
++do_work|code-block|syntax|javascript|3256693|无论您做什么，您都必须进行两次遍历(如果您的记录是固定长度的-这是不太可能的-您可以只获得文件大小并进行划分，但假设情况并非如此)。再次打开该文件确实不会花费您太多成本，但您可以避免这样做，如下所示。仅仅为了使用len()而转换为列表可能会浪费大量内存，而且速度也不会更快。|offset|length|style|CODE|3256694|注意：“Pythonic”的方法是使用enumerate而不是%2B=，但是UNPACK_TUPLE操作码的开销太大，使得enumerate比递增local更慢。话虽如此，这可能是一个不必要的微优化，您可能应该避免。|3256695|更多注释:如果你真的只想生成某种进度指示器，它不一定是基于记录的。您可以在循环中对文件对象执行tell()操作，并且只报告您正在处理的数据的百分比。这可能会有点不均匀，但任何足够大的文件都有可能需要一个进度条，记录长度的偏差将在噪音中消失。|3256696|entityMap^0|0|0|2Y|5|0|J|9|V|2|10|C|1N|9|0|1B|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]|$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

You only need to open the file once:

<pre><code>import csv

f = open('myFile.csv', 'rb')

countrdr = csv.DictReader(f)
totalrows = 0
for row in countrdr:
 totalrows += 1

f.seek(0) # You may not have to do this, I didn't check to see if DictReader did

myreader = csv.DictReader(f)
for row in myreader:
 do_work
</code></pre>

No matter what you do you have to make two passes (well, if your records are a fixed length - which is unlikely - you could just get the file size and divide, but lets presume that isn't the case). Opening the file again really doesn't cost you much, but you can avoid it as illustrated here. Converting to a list just to use <code>len()</code> is potentially going to waste tons of memory, and not be any faster.

Note: The 'Pythonic' way is to use <code>enumerate</code> instead of <code>+=</code>, but the <code>UNPACK_TUPLE</code> opcode is so expensive that it makes <code>enumerate</code> slower than incrementing a local. That being said, it's likely an unnecessary micro-optimization that you should probably avoid.

More Notes: If you really just want to generate some kind of progress indicator, it doesn't necessarily have to be record based. You can <code>tell()</code> on the file object in the loop and just report what % of the data you're through. It'll be a little uneven, but chances are on any file that's large enough to warrant a progress bar the deviation on record length will be lost in the noise.

blocks|key|3042800|text|正如答案https://stackoverflow.com/a/2890569/8056572中所提到的，您可以通过将阅读器的长度转换为列表来获得行数。然而，这将对RAM消耗产生影响，并且您将失去读取器(作为生成器)的好处。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3042801|在我看来，最好的解决方案是打开文件2次：|3042802|3042803|计算行数：|ordered-list-item|3042804|3042805|total_rows+=+sum(1+for+_+in+open('myFile.csv'))+#+-1+if+you+want+to+remove+the+header+from+the+count|code-block|syntax|javascript|3042806|注意:我使用.readlines()并不是为了避免加载内存中的所有行|style|CODE|3042807|3042808|遍历行|3042809|3042810|3042811|根据您的代码片段，您将得到如下内容：|3042812|import+csv

totalrows+=+sum(1+for+_+in+open('myFile.csv'))

myreader+=+csv.DictReader(open('myFile.csv'))

for+i,+_+in+enumerate(myreader,+start=1):
++++print("Row+%25d/%25d"+%25+(i,+totalrows))|3042813|注意：enumerate中的start=1表示i的第一个值。缺省情况下它是0，如果您保留这个缺省值，则必须在打印语句中使用i+%2B+1|3042814|3042815|如果您确实不想打开该文件两次，您可以使用答案https://stackoverflow.com/a/2891061/8056572中提到的seek|3042816|import+csv

f+=+open('myFile.csv')

total_rows+=+sum(1+for+_+in+f)

f.seek(0)

myreader+=+csv.DictReader(f)

for+i,+_+in+enumerate(myreader,+start=1):
++++print("Row+%25d/%25d"+%25+(i,+totalrows))|3042817|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2890569/8056572|1|https://stackoverflow.com/a/2891061/8056572^0|4|17|0|0|0|0|0|0|0|6|C|0|0|0|0|0|0|0|3|9|E|7|1P|5|0|0|1X|4|M|17|1|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@]|9|@$A|1K|B|1L|1|1M]]|C|$]]|$1|D|3|E|5|6|7|1N|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|1O|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|1P|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|1R|8|@]|9|@]|C|$N|O]]|$1|P|3|Q|5|6|7|1S|8|@$A|1T|B|1U|R|S]]|9|@]|C|$]]|$1|T|3|-4|5|6|7|1V|8|@]|9|@]|C|$]]|$1|U|3|V|5|I|7|1W|8|@]|9|@]|C|$]]|$1|W|3|-4|5|6|7|1X|8|@]|9|@]|C|$]]|$1|X|3|-4|5|6|7|1Y|8|@]|9|@]|C|$]]|$1|Y|3|Z|5|6|7|1Z|8|@]|9|@]|C|$]]|$1|10|3|11|5|M|7|20|8|@]|9|@]|C|$N|O]]|$1|12|3|13|5|6|7|21|8|@$A|22|B|23|R|S]|$A|24|B|25|R|S]|$A|26|B|27|R|S]]|9|@]|C|$]]|$1|14|3|-4|5|6|7|28|8|@]|9|@]|C|$]]|$1|15|3|16|5|6|7|29|8|@$A|2A|B|2B|R|S]]|9|@$A|2C|B|2D|1|2E]]|C|$]]|$1|17|3|18|5|M|7|2F|8|@]|9|@]|C|$N|O]]|$1|19|3|-4|5|6|7|2G|8|@]|9|@]|C|$]]]|1A|$1B|$5|1C|1D|1E|C|$1F|1G]]|1H|$5|1C|1D|1E|C|$1F|1I]]]]

As mentioned in the answer <a href="https://stackoverflow.com/a/2890569/8056572">https://stackoverflow.com/a/2890569/8056572</a> you can get the number of lines by taking the length of the reader converted to a list. However, this will have an impact on the RAM consumption and you will loose the benefits of the reader (which is a generator).
The best solution in my opinion is to open the file 2 times:
<ol>
<li>count the number of lines:</li>
</ol>
<pre class="lang-py prettyprint-override"><code>total_rows = sum(1 for _ in open('myFile.csv')) # -1 if you want to remove the header from the count
</code></pre>
Note: I am not using <code>.readlines()</code> to avoid to load all the lines in memory
<ol start="2">
<li>iterate over the lines</li>
</ol>
<hr />
According to your snippet you will have something like this:
<pre class="lang-py prettyprint-override"><code>import csv

totalrows = sum(1 for _ in open('myFile.csv'))

myreader = csv.DictReader(open('myFile.csv'))

for i, _ in enumerate(myreader, start=1):
 print(&quot;Row %d/%d&quot; % (i, totalrows))
</code></pre>
Note: the <code>start=1</code> in the <code>enumerate</code> indicates the first value of i. By default it is 0, if you keep this default value you have to use <code>i + 1</code> in the print statement
<hr />
If you really do not want to open the file two times you can use <code>seek</code> as mentioned in the answer <a href="https://stackoverflow.com/a/2891061/8056572">https://stackoverflow.com/a/2891061/8056572</a>
<pre class="lang-py prettyprint-override"><code>import csv

f = open('myFile.csv')

total_rows = sum(1 for _ in f)

f.seek(0)

myreader = csv.DictReader(f)

for i, _ in enumerate(myreader, start=1):
 print(&quot;Row %d/%d&quot; % (i, totalrows))
</code></pre>

I have a csv DictReader object (using Python 3.1), but I would like to know the number of lines/rows contained in the reader before I iterate through it. Something like as follows...

<pre><code>myreader = csv.DictReader(open('myFile.csv', newline=''))

totalrows = ?

rowcount = 0
for row in myreader:
 rowcount +=1
 print("Row %d/%d" % (rowcount,totalrows))
</code></pre>

I know I could get the total by iterating through the reader, but then I couldn't run the 'for' loop. I could iterate through a copy of the reader, but I cannot find how to copy an iterator. 

I could also use 

<pre><code>totalrows = len(open('myFile.csv').readlines())
</code></pre>

but that seems an unnecessary re-opening of the file. I would rather get the count from the DictReader if possible.

Any help would be appreciated.

Alan

Number of lines in csv.DictReader

我有一个csv DictReader对象(使用Python3.1)，但是在遍历它之前，我想知道阅读器中包含的行数/行数。类似于下面的内容。myreader = csv.DictReader(open('myFile.csv', newline=''))totalrows = ?rowcount = 0for row i...

问csv.DictReader中的行数
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