blocks|key|2813684|text|对于数据帧df，可以使用以下任一选项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2813685|2813686|len(df.index)|unordered-list-item|2813687|df.shape[0]|2813688|df[df.columns[0]].count()+(==+number+of+non-NaN+values+in+first+column)|2813689|2813690|​|2813691|📷|atomic|2813692|2813693|2813694|重现该图的代码：|2813695|import+numpy+as+np
import+pandas+as+pd
import+perfplot

perfplot.save(
++++"out.png",
++++setup=lambda+n:+pd.DataFrame(np.arange(n+*+3).reshape(n,+3)),
++++n_range=[2**k+for+k+in+range(25)],
++++kernels=[
++++++++lambda+df:+len(df.index),
++++++++lambda+df:+df.shape[0],
++++++++lambda+df:+df[df.columns[0]].count(),
++++],
++++labels=["len(df.index)",+"df.shape[0]",+"df[df.columns[0]].count()"],
++++xlabel="Number+of+rows",
)|code-block|syntax|javascript|2813696|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/docs/reference/api/pandas.Series.count.html|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/6e5d44f4a26c238d5d2986aba353ed68.png|imageAlt^0|5|2|0|0|0|D|0|0|B|0|0|P|U|O|0|0|0|0|0|1|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|J|3|K|5|I|7|1N|8|@$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|L|3|M|5|I|7|1Q|8|@$9|1R|A|1S|B|C]]|D|@$9|1T|A|1U|1|1V]]|E|$]]|$1|N|3|-4|5|6|7|1W|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1X|8|@]|D|@]|E|$]]|$1|Q|3|R|5|S|7|1Y|8|@]|D|@$9|1Z|A|20|1|21]]|E|$]]|$1|T|3|P|5|6|7|22|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|23|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|24|8|@]|D|@]|E|$]]|$1|X|3|Y|5|Z|7|25|8|@]|D|@]|E|$10|11]]|$1|12|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|13|$14|$5|15|16|17|E|$18|19]]|1A|$5|1B|16|1C|E|$1D|1E|1F|-4]]]]

For a dataframe <code>df</code>, one can use any of the following:
<ul>
<li><code>len(df.index)</code></li>
<li><code>df.shape[0]</code></li>
<li><code>df[df.columns[0]].count()</code> (== <a href="https://pandas.pydata.org/docs/reference/api/pandas.Series.count.html" rel="noreferrer">number of non-NaN values</a> in first column)</li>
</ul>
<a href="https://i.stack.imgur.com/wEzue.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wEzue.png" alt="Performance plot" /></a>
<hr />
Code to reproduce the plot:
<pre class="lang-py prettyprint-override"><code>import numpy as np
import pandas as pd
import perfplot

perfplot.save(
 &quot;out.png&quot;,
 setup=lambda n: pd.DataFrame(np.arange(n * 3).reshape(n, 3)),
 n_range=[2**k for k in range(25)],
 kernels=[
 lambda df: len(df.index),
 lambda df: df.shape[0],
 lambda df: df[df.columns[0]].count(),
 ],
 labels=[&quot;len(df.index)&quot;, &quot;df.shape[0]&quot;, &quot;df[df.columns[0]].count()&quot;],
 xlabel=&quot;Number of rows&quot;,
)
</code></pre>

blocks|key|2813789|text|使用len(df)+:-)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2813790|__len__()使用"Returns+length+of+index“进行记录。|2813791|计时信息，设置方式与root's+answer相同|2813792|In+[7]:+timeit+len(df.index)
1000000+loops,+best+of+3:+248+ns+per+loop

In+[8]:+timeit+len(df)
1000000+loops,+best+of+3:+573+ns+per+loop|code-block|syntax|javascript|2813793|由于有一个额外的函数调用，当然可以说它比直接调用len(df.index)慢一点。但在大多数情况下，这应该无关紧要。我发现len(df)非常具有可读性。|2813794|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/15943975#15943975^0|2|7|0|0|9|0|A|D|0|0|0|O|D|1P|7|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|J|3|K|5|L|7|18|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

Use <code>len(df)</code> :-).
<code>__len__()</code> is documented with &quot;Returns length of index&quot;.
Timing info, set up the same way as in <a href="https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/15943975#15943975">root's answer</a>:
<pre class="lang-none prettyprint-override"><code>In [7]: timeit len(df.index)
1000000 loops, best of 3: 248 ns per loop

In [8]: timeit len(df)
1000000 loops, best of 3: 573 ns per loop
</code></pre>
Due to one additional function call, it is of course correct to say that it is a bit slower than calling <code>len(df.index)</code> directly. But this should not matter in most cases. I find <code>len(df)</code> to be quite readable.

blocks|key|2813891|text|除了前面的答案之外，您还可以使用df.axes获取具有行和列索引的元组，然后使用len()函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2813892|total_rows+=+len(df.axes[0])
total_cols+=+len(df.axes[1])|code-block|syntax|javascript|2813893|entityMap^0|G|7|14|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Apart from the previous answers, you can use <code>df.axes</code> to get the tuple with row and column indexes and then use the <code>len()</code> function:
<pre><code>total_rows = len(df.axes[0])
total_cols = len(df.axes[1])
</code></pre>

blocks|key|191458|text|假设df是你的数据帧，那么：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|191459|count_row+=+df.shape[0]++#+Gives+number+of+rows
count_col+=+df.shape[1]++#+Gives+number+of+columns|code-block|syntax|javascript|191460|或者，更简洁地说，|191461|r,+c+=+df.shape|191462|entityMap^0|2|2|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|$]]

Suppose <code>df</code> is your dataframe then:
<pre><code>count_row = df.shape[0] # Gives number of rows
count_col = df.shape[1] # Gives number of columns
</code></pre>
Or, more succinctly,
<pre><code>r, c = df.shape
</code></pre>

blocks|key|3026475|text|使用+len(df)的TL;DR+|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|3026476|len()允许您获取列表中的项数。因此，要获得DataFrame的行数，只需使用len(df)。|3026477|或者，您可以分别使用df.index和df.columns访问所有行和所有列。由于您可以使用len(anyList)获取元素编号，因此使用len(df.index)将提供行数，使用len(df.columns)将提供列数。|3026478|或者，您可以使用df.shape，它返回行数和列数(作为一个元组)。如果您想访问行数，只需使用df.shape[0]。对于列数，仅使用：df.shape[1]。|3026479|entityMap^0|3|7|3|7|0|0|1C|0|5|14|7|0|0|33|A|8|J|A|1A|C|1X|D|2I|F|0|0|28|8|8|1B|B|1W|B|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|D]|$9|Y|A|Z|B|D]]|E|@]|F|$]]|$1|I|3|J|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|D]|$9|15|A|16|B|D]|$9|17|A|18|B|D]|$9|19|A|1A|B|D]|$9|1B|A|1C|B|D]]|E|@]|F|$]]|$1|K|3|L|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|D]|$9|1I|A|1J|B|D]|$9|1K|A|1L|B|D]]|E|@]|F|$]]|$1|M|3|-4|5|6|7|1M|8|@]|E|@]|F|$]]]|N|$]]

<h2>TL;DR use <code>len(df)</code></h2>
<code>len()</code> lets you for getting the number of items in a list. So, for getting row counts of a DataFrame, simply use <code>len(df)</code>.
Alternatively, you can access all rows and all columns with <code>df.index</code>, and <code>df.columns</code>,respectively. Since you can use the <code>len(anyList)</code> for getting the element numbers, use
<code>len(df.index)</code> will give you the number of rows, and <code>len(df.columns)</code> will give the number of columns.
Or, you can use <code>df.shape</code> which returns the number of rows and columns together (as a tuple). If you want to access the number of rows, only use <code>df.shape[0]</code>. For the number of columns, only use: <code>df.shape[1]</code>.

blocks|key|2813933|text|我是从R背景来到Pandas的，我看到Pandas在选择行或列时更加复杂。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2813934|我不得不和它搏斗了一段时间，然后我找到了一些方法来处理：|2813935|获取列数：|2813936|len(df.columns)
##+Here:
#+df+is+your+data.frame
#+df.columns+returns+a+string.+It+contains+column's+titles+of+the+df.
#+Then,+"len()"+gets+the+length+of+it.|code-block|syntax|javascript|2813937|获取行数：|2813938|len(df.index)+#+It's+similar.|2813939|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/R_%2528programming_language%2529^0|3|1|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|16|8|@]|9|@]|C|$K|L]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

I come to Pandas from an <a href="https://en.wikipedia.org/wiki/R_%28programming_language%29" rel="nofollow noreferrer">R</a> background, and I see that Pandas is more complicated when it comes to selecting rows or columns.
I had to wrestle with it for a while, and then I found some ways to deal with:
Getting the number of columns:
<pre><code>len(df.columns)
## Here:
# df is your data.frame
# df.columns returns a string. It contains column's titles of the df.
# Then, &quot;len()&quot; gets the length of it.
</code></pre>
Getting the number of rows:
<pre><code>len(df.index) # It's similar.
</code></pre>

blocks|key|3026502|text|对于dataframe+df，是在浏览数据时使用的打印逗号格式的行数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3026503|def+nrow(df):
++++print("{:,}".format(df.shape[0]))|code-block|syntax|javascript|3026504|示例：|3026505|nrow(my_df)
12,456,789|3026506|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|$]]

For dataframe df, a printed comma formatted row count used while exploring data: 

<pre><code>def nrow(df):
 print("{:,}".format(df.shape[0]))
</code></pre>

Example:

<pre><code>nrow(my_df)
12,456,789
</code></pre>

blocks|key|193898|text|Jan-Philip+Gehrcke's+answer上的...building。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|193899|len(df)或len(df.index)比df.shape[0]快的原因|style|CODE|193900|看看代码。df.shape是一个@property，它运行一个调用len两次的DataFrame方法。|193901|df.shape??
Type:++++++++property
String+form:+<property+object+at+0x1127b33c0>
Source:
#+df.shape.fget
@property
def+shape(self):
++++"""
++++Return+a+tuple+representing+the+dimensionality+of+the+DataFrame.
++++"""
++++return+len(self.index),+len(self.columns)|code-block|syntax|javascript|193902|在len(df)的引擎盖下|193903|df.__len__??
Signature:+df.__len__()
Source:
++++def+__len__(self):
++++++++"""Returns+length+of+info+axis,+but+here+we+use+the+index+"""
++++++++return+len(self.index)
File:++++++~/miniconda2/lib/python2.7/site-packages/pandas/core/frame.py
Type:++++++instancemethod|193904|len(df.index)会比len(df)稍微快一点，因为它的函数调用少了一次，但这总是比df.shape[0]快|193905|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/18317067#18317067^0|0|R|0|0|0|7|8|D|M|B|0|G|9|X|3|0|0|0|0|0|D|F|7|1A|B|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|6|7|16|8|@$A|17|B|18|F|G]|$A|19|B|1A|F|G]|$A|1B|B|1C|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|1D|8|@$A|1E|B|1F|F|G]|$A|1G|B|1H|F|G]]|9|@]|C|$]]|$1|J|3|K|5|L|7|1I|8|@]|9|@]|C|$M|N]]|$1|O|3|P|5|6|7|1J|8|@]|9|@]|C|$]]|$1|Q|3|R|5|L|7|1K|8|@]|9|@]|C|$M|N]]|$1|S|3|T|5|6|7|1L|8|@$A|1M|B|1N|F|G]|$A|1O|B|1P|F|G]|$A|1Q|B|1R|F|G]]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1S|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

...building on <a href="https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/18317067#18317067">Jan-Philip Gehrcke's answer</a>.
The reason why <code>len(df)</code> or <code>len(df.index)</code> is faster than <code>df.shape[0]</code>:
Look at the code. df.shape is a <code>@property</code> that runs a DataFrame method calling <code>len</code> twice.
<pre><code>df.shape??
Type: property
String form: &lt;property object at 0x1127b33c0&gt;
Source:
# df.shape.fget
@property
def shape(self):
 &quot;&quot;&quot;
 Return a tuple representing the dimensionality of the DataFrame.
 &quot;&quot;&quot;
 return len(self.index), len(self.columns)
</code></pre>
And beneath the hood of len(df)
<pre><code>df.__len__??
Signature: df.__len__()
Source:
 def __len__(self):
 &quot;&quot;&quot;Returns length of info axis, but here we use the index &quot;&quot;&quot;
 return len(self.index)
File: ~/miniconda2/lib/python2.7/site-packages/pandas/core/frame.py
Type: instancemethod
</code></pre>
<code>len(df.index)</code> will be slightly faster than <code>len(df)</code> since it has one less function call, but this is always faster than <code>df.shape[0]</code>

blocks|key|193907|text|如果您希望在链接操作过程中获得行数，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193908|df.pipe(len)|code-block|syntax|javascript|193909|示例：|193910|row_count+=+(
++++++pd.DataFrame(np.random.rand(3,4))
++++++.reset_index()
++++++.pipe(len)
)|193911|如果您不想将长语句放入len()函数中，这将非常有用。|offset|length|style|CODE|193912|你可以使用__len__()，但是__len__()看起来有点奇怪。|193913|entityMap^0|0|0|0|0|B|5|0|5|9|H|9|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@$M|12|N|13|O|P]|$M|14|N|15|O|P]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|T|$]]

In case you want to get the row count in the middle of a chained operation, you can use:

<pre><code>df.pipe(len)
</code></pre>

Example:

<pre><code>row_count = (
 pd.DataFrame(np.random.rand(3,4))
 .reset_index()
 .pipe(len)
)
</code></pre>

This can be useful if you don't want to put a long statement inside a <code>len()</code> function.

You could use <code>__len__()</code> instead but <code>__len__()</code> looks a bit weird.

blocks|key|191860|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|191861|如何获取熊猫DataFrame的行数？|blockquote|191862|191863|191864|此表总结了您想要计算DataFrame+(或系列，以确保完整性)中的某些内容的不同情况，以及推荐的方法。|191865|​|191866|📷|atomic|offset|length|191867|191868|191869|+脚注|style|BOLD|191870|191871|191872|191873|DataFrame.count返回每列的计数作为Series，因为非null计数因非null计数而异。column.|ordered-list-item|CODE|191874|DataFrameGroupBy.size返回Series，因为同一组中的所有列共享相同的row-count.|191875|DataFrameGroupBy.count返回DataFrame，因为同一组中的不同列的非null计数可能不同。若要获取特定列的分组非空计数，请使用df.groupby(...)['x'].count()，其中"x“是要计数的列。|191876|191877|191878|#最小代码示例|191879|下面，我展示了上表中描述的每种方法的示例。首先，设置-|191880|df+=+pd.DataFrame({
++++'A':+list('aabbc'),+'B':+['x',+'x',+np.nan,+'x',+np.nan]})
s+=+df['B'].copy()

df

+++A++++B
0++a++++x
1++a++++x
2++b++NaN
3++b++++x
4++c++NaN

s

0++++++x
1++++++x
2++++NaN
3++++++x
4++++NaN
Name:+B,+dtype:+object|code-block|syntax|javascript|191881|DataFrame的行数：len(df)、df.shape[0]或len(df.index)|191882|len(df)
#+5

df.shape[0]
#+5

len(df.index)
#+5|191883|比较固定时间操作的性能似乎很愚蠢，特别是当差异是在“认真的，不用担心”的水平上时。但这似乎是其他答案的一种趋势，所以为了完整性，我也这么做了。|191884|在上述三种方法中，len(df.index)+(如其他答案中提到的)是最快的。|191885|191886|+备注|191887|191888|191889|以上所有方法都是常量时间操作，因为它们都很简单，属性lookups.|191890|df.shape+(类似于|unordered-list-item|191891|)是一个返回(#+Rows,+#+Cols)的元组的属性。例如，对于此处的示例，df.shape返回(8,+2)。|191892|191893|191894|DataFrame的列数：df.shape[1]，len(df.columns)|191895|df.shape[1]
#+2

len(df.columns)
#+2|191896|与len(df.index)类似，len(df.columns)是两种方法中速度更快的一种(但需要输入更多字符)。|191897|系列的行数：len(s)、s.size、len(s.index)|191898|len(s)
#+5

s.size
#+5

len(s.index)
#+5|191899|就速度而言，s.size和len(s.index)大致相同。但我推荐使用len(df)。|191900|191901|++size是一个属性，它返回元素的数量(=任意系列的行数)。DataFrames还定义了一个size属性，它返回与df.shape[0]+*+df.shape[1]相同的结果。|191902|191903|191904|非空行计数：DataFrame.count和Series.count|191905|这里描述的方法只计算非空值(这意味着忽略NaNs+)。|191906|调用DataFrame.count将返回每列的非NaN计数：|191907|df.count()

A++++5
B++++3
dtype:+int64|191908|对于系列，使用Series.count可以达到类似的效果：|191909|s.count()
#+3|191910|分组行数：GroupBy.size|191911|对于DataFrames，使用DataFrameGroupBy.size计算每个组的行数。|191912|df.groupby('A').size()

A
a++++2
b++++2
c++++1
dtype:+int64|191913|类似地，对于Series，您将使用SeriesGroupBy.size。|191914|s.groupby(df.A).size()

A
a++++2
b++++2
c++++1
Name:+B,+dtype:+int64|191915|在这两种情况下，都会返回一个Series。这对于DataFrames也是有意义的，因为所有组共享相同的行数。|191916|按组计算的非空行计数：GroupBy.count|191917|与上面类似，但使用GroupBy.count，而不是GroupBy.size。请注意，size始终返回一个Series，而如果对特定列调用，count将返回一个Series，否则返回一个DataFrame。|191918|下面的方法返回相同的东西：|191919|df.groupby('A')['B'].size()
df.groupby('A').size()

A
a++++2
b++++2
c++++1
Name:+B,+dtype:+int64|191920|同时，对于count，我们有|191921|df.groupby('A').count()

+++B
A
a++2
b++1
c++0|191922|整个GroupBy对象的...called，而不是|191923|df.groupby('A')['B'].count()

A
a++++2
b++++1
c++++0
Name:+B,+dtype:+int64|191924|在特定列上调用。|191925|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/f65a38c897877cb257f155e94e164738.png|imageAlt|1|LINK|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.count.html|2|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.count.html|3|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.DataFrameGroupBy.size.html|4|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.GroupBy.size.html|5^0|0|0|0|0|0|0|0|1|0|0|0|0|1|2|0|0|0|0|0|F|O|6|0|0|L|N|6|0|0|M|O|9|23|S|0|0|0|1|6|0|0|0|D|7|L|B|X|D|0|0|0|9|D|0|0|1|2|0|0|0|0|0|8|0|6|G|14|8|1E|6|0|0|0|D|B|P|F|0|0|1|D|H|F|0|6|6|D|6|K|C|0|0|6|6|D|C|10|7|0|0|2|4|1M|P|0|0|0|6|F|M|C|0|0|2|F|2|F|1|0|0|7|C|7|C|2|0|0|5|C|0|2|A|F|L|F|L|3|0|0|6|6|H|I|H|I|4|0|0|E|6|O|A|0|B|D|0|9|D|Q|C|17|4|1H|6|1Y|5|28|6|2L|9|9|D|5|0|0|0|5|5|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|4A|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|4B|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|4C|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|4D|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|4E|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|4F|7|@]|8|@]|9|$]]|$1|J|3|K|4|L|6|4G|7|@]|8|@$M|4H|N|4I|1|4J]]|9|$]]|$1|O|3|I|4|5|6|4K|7|@]|8|@]|9|$]]|$1|P|3|-4|4|5|6|4L|7|@]|8|@]|9|$]]|$1|Q|3|R|4|C|6|4M|7|@$M|4N|N|4O|S|T]]|8|@]|9|$]]|$1|U|3|-4|4|5|6|4P|7|@]|8|@]|9|$]]|$1|V|3|-4|4|5|6|4Q|7|@]|8|@]|9|$]]|$1|W|3|-4|4|5|6|4R|7|@]|8|@]|9|$]]|$1|X|3|Y|4|Z|6|4S|7|@$M|4T|N|4U|S|10]|$M|4V|N|4W|S|10]]|8|@]|9|$]]|$1|11|3|12|4|Z|6|4X|7|@$M|4Y|N|4Z|S|10]|$M|50|N|51|S|10]]|8|@]|9|$]]|$1|13|3|14|4|Z|6|52|7|@$M|53|N|54|S|10]|$M|55|N|56|S|10]|$M|57|N|58|S|10]]|8|@]|9|$]]|$1|15|3|-4|4|5|6|59|7|@]|8|@]|9|$]]|$1|16|3|-4|4|5|6|5A|7|@]|8|@]|9|$]]|$1|17|3|18|4|5|6|5B|7|@$M|5C|N|5D|S|T]]|8|@]|9|$]]|$1|19|3|1A|4|5|6|5E|7|@]|8|@]|9|$]]|$1|1B|3|1C|4|1D|6|5F|7|@]|8|@]|9|$1E|1F]]|$1|1G|3|1H|4|5|6|5G|7|@$M|5H|N|5I|S|10]|$M|5J|N|5K|S|10]|$M|5L|N|5M|S|10]]|8|@]|9|$]]|$1|1I|3|1J|4|1D|6|5N|7|@]|8|@]|9|$1E|1F]]|$1|1K|3|1L|4|5|6|5O|7|@]|8|@]|9|$]]|$1|1M|3|1N|4|5|6|5P|7|@$M|5Q|N|5R|S|10]]|8|@]|9|$]]|$1|1O|3|-4|4|5|6|5S|7|@]|8|@]|9|$]]|$1|1P|3|1Q|4|C|6|5T|7|@$M|5U|N|5V|S|T]]|8|@]|9|$]]|$1|1R|3|-4|4|5|6|5W|7|@]|8|@]|9|$]]|$1|1S|3|-4|4|5|6|5X|7|@]|8|@]|9|$]]|$1|1T|3|1U|4|5|6|5Y|7|@]|8|@]|9|$]]|$1|1V|3|1W|4|1X|6|5Z|7|@$M|60|N|61|S|10]]|8|@]|9|$]]|$1|1Y|3|1Z|4|1X|6|62|7|@$M|63|N|64|S|10]|$M|65|N|66|S|10]|$M|67|N|68|S|10]]|8|@]|9|$]]|$1|20|3|-4|4|5|6|69|7|@]|8|@]|9|$]]|$1|21|3|-4|4|5|6|6A|7|@]|8|@]|9|$]]|$1|22|3|23|4|5|6|6B|7|@$M|6C|N|6D|S|10]|$M|6E|N|6F|S|10]]|8|@]|9|$]]|$1|24|3|25|4|1D|6|6G|7|@]|8|@]|9|$1E|1F]]|$1|26|3|27|4|5|6|6H|7|@$M|6I|N|6J|S|10]|$M|6K|N|6L|S|10]]|8|@]|9|$]]|$1|28|3|29|4|5|6|6M|7|@$M|6N|N|6O|S|10]|$M|6P|N|6Q|S|10]|$M|6R|N|6S|S|10]]|8|@]|9|$]]|$1|2A|3|2B|4|1D|6|6T|7|@]|8|@]|9|$1E|1F]]|$1|2C|3|2D|4|5|6|6U|7|@$M|6V|N|6W|S|10]|$M|6X|N|6Y|S|10]|$M|6Z|N|70|S|10]]|8|@]|9|$]]|$1|2E|3|-4|4|5|6|71|7|@]|8|@]|9|$]]|$1|2F|3|2G|4|C|6|72|7|@$M|73|N|74|S|10]|$M|75|N|76|S|10]]|8|@]|9|$]]|$1|2H|3|-4|4|5|6|77|7|@]|8|@]|9|$]]|$1|2I|3|-4|4|5|6|78|7|@]|8|@]|9|$]]|$1|2J|3|2K|4|5|6|79|7|@$M|7A|N|7B|S|10]|$M|7C|N|7D|S|10]]|8|@]|9|$]]|$1|2L|3|2M|4|5|6|7E|7|@]|8|@]|9|$]]|$1|2N|3|2O|4|5|6|7F|7|@$M|7G|N|7H|S|10]]|8|@$M|7I|N|7J|1|7K]]|9|$]]|$1|2P|3|2Q|4|1D|6|7L|7|@]|8|@]|9|$1E|1F]]|$1|2R|3|2S|4|5|6|7M|7|@$M|7N|N|7O|S|10]]|8|@$M|7P|N|7Q|1|7R]]|9|$]]|$1|2T|3|2U|4|1D|6|7S|7|@]|8|@]|9|$1E|1F]]|$1|2V|3|2W|4|5|6|7T|7|@$M|7U|N|7V|S|10]]|8|@]|9|$]]|$1|2X|3|2Y|4|5|6|7W|7|@$M|7X|N|7Y|S|10]|$M|7Z|N|80|S|10]]|8|@$M|81|N|82|1|83]]|9|$]]|$1|2Z|3|30|4|1D|6|84|7|@]|8|@]|9|$1E|1F]]|$1|31|3|32|4|5|6|85|7|@$M|86|N|87|S|10]|$M|88|N|89|S|10]]|8|@$M|8A|N|8B|1|8C]]|9|$]]|$1|33|3|34|4|1D|6|8D|7|@]|8|@]|9|$1E|1F]]|$1|35|3|36|4|5|6|8E|7|@$M|8F|N|8G|S|10]|$M|8H|N|8I|S|10]]|8|@]|9|$]]|$1|37|3|38|4|5|6|8J|7|@$M|8K|N|8L|S|10]]|8|@]|9|$]]|$1|39|3|3A|4|5|6|8M|7|@$M|8N|N|8O|S|10]|$M|8P|N|8Q|S|10]|$M|8R|N|8S|S|10]|$M|8T|N|8U|S|10]|$M|8V|N|8W|S|10]|$M|8X|N|8Y|S|10]|$M|8Z|N|90|S|10]]|8|@$M|91|N|92|1|93]]|9|$]]|$1|3B|3|3C|4|5|6|94|7|@]|8|@]|9|$]]|$1|3D|3|3E|4|1D|6|95|7|@]|8|@]|9|$1E|1F]]|$1|3F|3|3G|4|5|6|96|7|@$M|97|N|98|S|10]]|8|@]|9|$]]|$1|3H|3|3I|4|1D|6|99|7|@]|8|@]|9|$1E|1F]]|$1|3J|3|3K|4|5|6|9A|7|@]|8|@]|9|$]]|$1|3L|3|3M|4|1D|6|9B|7|@]|8|@]|9|$1E|1F]]|$1|3N|3|3O|4|5|6|9C|7|@]|8|@]|9|$]]|$1|3P|3|-4|4|5|6|9D|7|@]|8|@]|9|$]]]|3Q|$3R|$4|3S|3T|3U|9|$3V|3W|3X|-4]]|3Y|$4|3Z|3T|40|9|$41|42]]|43|$4|3Z|3T|40|9|$41|44]]|45|$4|3Z|3T|40|9|$41|46]]|47|$4|3Z|3T|40|9|$41|48]]|49|$4|3Z|3T|40|9|$41|48]]]]

<blockquote>
<h1>How do I get the row count of a Pandas DataFrame?</h1>
</blockquote>
This table summarises the different situations in which you'd want to count something in a DataFrame (or Series, for completeness), along with the recommended method(s).
<a href="https://i.stack.imgur.com/3FXuI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3FXuI.png" alt="Enter image description here" /></a>
<blockquote>
Footnotes
<ol>
<li><code>DataFrame.count</code> returns counts for each column as a <code>Series</code> since the non-null count varies by column.</li>
<li><code>DataFrameGroupBy.size</code> returns a <code>Series</code>, since all columns in the same group share the same row-count.</li>
<li><code>DataFrameGroupBy.count</code> returns a <code>DataFrame</code>, since the non-null count could differ across columns in the same group. To get the group-wise non-null count for a specific column, use <code>df.groupby(...)['x'].count()</code> where &quot;x&quot; is the column to count.</li>
</ol>
</blockquote>
<hr />
#Minimal Code Examples
Below, I show examples of each of the methods described in the table above. First, the setup -
<pre><code>df = pd.DataFrame({
 'A': list('aabbc'), 'B': ['x', 'x', np.nan, 'x', np.nan]})
s = df['B'].copy()

df

 A B
0 a x
1 a x
2 b NaN
3 b x
4 c NaN

s

0 x
1 x
2 NaN
3 x
4 NaN
Name: B, dtype: object
</code></pre>
<h3>Row Count of a DataFrame: <code>len(df)</code>, <code>df.shape[0]</code>, or <code>len(df.index)</code></h3>
<pre><code>len(df)
# 5

df.shape[0]
# 5

len(df.index)
# 5
</code></pre>
It seems silly to compare the performance of constant time operations, especially when the difference is on the level of &quot;seriously, don't worry about it&quot;. But this seems to be a trend with other answers, so I'm doing the same for completeness.
Of the three methods above, <code>len(df.index)</code> (as mentioned in other answers) is the fastest.
<blockquote>
Note
<ul>
<li>All the methods above are constant time operations as they are simple attribute lookups.</li>
<li><code>df.shape</code> (similar to <code>ndarray.shape</code>) is an attribute that returns a tuple of <code>(# Rows, # Cols)</code>. For example, <code>df.shape</code> returns <code>(8, 2)</code> for the example here.</li>
</ul>
</blockquote>
<h3>Column Count of a DataFrame: <code>df.shape[1]</code>, <code>len(df.columns)</code></h3>
<pre><code>df.shape[1]
# 2

len(df.columns)
# 2
</code></pre>
Analogous to <code>len(df.index)</code>, <code>len(df.columns)</code> is the faster of the two methods (but takes more characters to type).
<h3>Row Count of a Series: <code>len(s)</code>, <code>s.size</code>, <code>len(s.index)</code></h3>
<pre><code>len(s)
# 5

s.size
# 5

len(s.index)
# 5
</code></pre>
<code>s.size</code> and <code>len(s.index)</code> are about the same in terms of speed. But I recommend <code>len(df)</code>.
<blockquote>
Note
<code>size</code> is an attribute, and it returns the number of elements (=count
of rows for any Series). DataFrames also define a size attribute which
returns the same result as <code>df.shape[0] * df.shape[1]</code>.
</blockquote>
<h3>Non-Null Row Count: <code>DataFrame.count</code> and <code>Series.count</code></h3>
The methods described here only count non-null values (meaning NaNs are ignored).
Calling <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.count.html" rel="noreferrer"><code>DataFrame.count</code></a> will return non-NaN counts for each column:
<pre><code>df.count()

A 5
B 3
dtype: int64
</code></pre>
For Series, use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.count.html" rel="noreferrer"><code>Series.count</code></a> to similar effect:
<pre><code>s.count()
# 3
</code></pre>
<h3>Group-wise Row Count: <code>GroupBy.size</code></h3>
For <code>DataFrames</code>, use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.DataFrameGroupBy.size.html" rel="noreferrer"><code>DataFrameGroupBy.size</code></a> to count the number of rows per group.
<pre><code>df.groupby('A').size()

A
a 2
b 2
c 1
dtype: int64
</code></pre>
Similarly, for <code>Series</code>, you'll use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.GroupBy.size.html" rel="noreferrer"><code>SeriesGroupBy.size</code></a>.
<pre><code>s.groupby(df.A).size()

A
a 2
b 2
c 1
Name: B, dtype: int64
</code></pre>
In both cases, a <code>Series</code> is returned. This makes sense for <code>DataFrames</code> as well since all groups share the same row-count.
<h3>Group-wise Non-Null Row Count: <code>GroupBy.count</code></h3>
Similar to above, but use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.GroupBy.size.html" rel="noreferrer"><code>GroupBy.count</code></a>, not <code>GroupBy.size</code>. Note that <code>size</code> always returns a <code>Series</code>, while <code>count</code> returns a <code>Series</code> if called on a specific column, or else a <code>DataFrame</code>.
The following methods return the same thing:
<pre><code>df.groupby('A')['B'].size()
df.groupby('A').size()

A
a 2
b 2
c 1
Name: B, dtype: int64
</code></pre>
Meanwhile, for <code>count</code>, we have
<pre><code>df.groupby('A').count()

 B
A
a 2
b 1
c 0
</code></pre>
...called on the entire GroupBy object, vs.,
<pre><code>df.groupby('A')['B'].count()

A
a 2
b 1
c 0
Name: B, dtype: int64
</code></pre>
Called on a specific column.

blocks|key|193934|text|要找出数据帧中的行数，另一种方法是pandas.Index.size，我认为这是最易读的变体。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|193935|请注意，正如我在the+accepted+answer上所评论的那样，|193936|193937|怀疑pandas.Index.size实际上会比len(df.index)更快，但我电脑上的timeit告诉我并非如此(每个循环大约慢150+ns+)。|blockquote|193938|193939|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Index.size.html|1|https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/15943975#15943975^0|H|H|H|H|0|0|8|J|1|0|0|2|H|O|D|1A|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|H|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

An alternative method to finding out the amount of rows in a dataframe which I think is the most readable variant is <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Index.size.html" rel="nofollow noreferrer"><code>pandas.Index.size</code></a>.
Do note that, as I commented on <a href="https://stackoverflow.com/questions/15943769/how-do-i-get-the-row-count-of-a-pandas-dataframe/15943975#15943975">the accepted answer</a>,
<blockquote>
Suspected <code>pandas.Index.size</code> would actually be faster than <code>len(df.index)</code> but <code>timeit</code> on my computer tells me otherwise (~150 ns slower per loop).
</blockquote>

blocks|key|193941|text|我不确定这是否会起作用(可以省略数据)，但这可能会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193942|*dataframe+name*.tails(1)|code-block|syntax|javascript|193943|然后使用它，您可以通过运行代码片段并查看提供给您的行号来找到行数。|193944|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|$]]

I'm not sure if this would work (data could be omitted), but this may work:
<pre><code>*dataframe name*.tails(1)
</code></pre>
and then using this, you could find the number of rows by running the code snippet and looking at the row number that was given to you.

blocks|key|2814041|text|这两种方法都可以做到(df是DataFrame的名称)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2814042|方法1：使用len函数：|BOLD|2814043|len(df)将给出名为df的DataFrame中的行数。|2814044|方法2：使用count函数：|2814045|df[col].count()将计算给定列col中的行数。|2814046|df.count()将给出所有列的行数。|2814047|entityMap^0|B|2|0|0|3|6|3|0|0|7|C|2|0|0|3|6|5|0|0|F|L|3|0|0|A|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|H]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|I|3|J|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|H]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|R|$]]

Either of this can do it (<code>df</code> is the name of the DataFrame):
Method 1: Using the <code>len</code> function:
<code>len(df)</code> will give the number of rows in a DataFrame named <code>df</code>.
Method 2: using <code>count</code> function:
<code>df[col].count()</code> will count the number of rows in a given column <code>col</code>.
<code>df.count()</code> will give the number of rows for all the columns.

blocks|key|3026665|text|您还可以执行此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3026666|假设df是您的数据帧。然后，df.shape为您提供数据帧的形状，即(row,col)|offset|length|style|CODE|3026667|因此，指定以下命令以获取所需的|3026668|+row+=+df.shape[0],+col+=+df.shape[1]|code-block|syntax|javascript|3026669|entityMap^0|0|2|2|E|8|Y|9|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

You can do this also:
Let’s say <code>df</code> is your dataframe. Then <code>df.shape</code> gives you the shape of your dataframe i.e <code>(row,col)</code>
Thus, assign the below command to get the required
<pre><code> row = df.shape[0], col = df.shape[1]
</code></pre>

blocks|key|2814065|text|假设数据集是"data“，将数据集命名为”data_fr“，data_fr中的行数为"nu_rows”。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2814066|#import+the+data+frame.+Extention+could+be+different+as+csv,xlsx+or+etc.
data_fr+=+pd.read_csv('data.csv')

#print+the+number+of+rows
nu_rows+=+data_fr.shape[0]
print(nu_rows)|code-block|syntax|javascript|2814067|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|$]]

Think, the dataset is &quot;data&quot; and name your dataset as &quot; data_fr &quot; and number of rows in the data_fr is &quot;nu_rows&quot;
<pre><code>#import the data frame. Extention could be different as csv,xlsx or etc.
data_fr = pd.read_csv('data.csv')

#print the number of rows
nu_rows = data_fr.shape[0]
print(nu_rows)
</code></pre>

blocks|key|193957|text|#+this+will+give+you+the+number+of+rows+in+the+dataframe+df
df.shape[0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|193958|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># this will give you the number of rows in the dataframe df
df.shape[0]
</code></pre>

I'm trying to get the number of rows of dataframe df with Pandas, and here is my code.
<h3>Method 1:</h3>
<pre><code>total_rows = df.count
print total_rows + 1
</code></pre>
<h3>Method 2:</h3>
<pre><code>total_rows = df['First_column_label'].count
print total_rows + 1
</code></pre>
Both the code snippets give me this error:
<blockquote>
TypeError: unsupported operand type(s) for +: 'instancemethod' and 'int'
</blockquote>
What am I doing wrong?

How do I get the row count of a Pandas DataFrame?

我正在尝试使用Pandas获取数据帧df的行数，下面是我的代码。方法1：total_rows = df.countprint total_rows + 1方法二：total_rows = df['First_column_label'].countprint total_rows + 1这两个代码片段都给出了这个错误：...

问如何获取熊猫DataFrame的行数？
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