blocks|key|190269|text|如果您的DataFrame只有一行，则使用iloc将第一(唯一)行作为系列访问，然后使用列名访问该值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|190270|In+[3]:+sub_df
Out[3]:
++++++++++A+++++++++B
2+-0.133653+-0.030854

In+[4]:+sub_df.iloc[0]
Out[4]:
A+++-0.133653
B+++-0.030854
Name:+2,+dtype:+float64

In+[5]:+sub_df.iloc[0]['A']
Out[5]:+-0.13365288513107493|code-block|syntax|javascript|190271|entityMap^0|L|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you have a DataFrame with only one row, then access the first (only) row as a Series using <code>iloc</code>, and then the value using the column name:

<pre><code>In [3]: sub_df
Out[3]:
 A B
2 -0.133653 -0.030854

In [4]: sub_df.iloc[0]
Out[4]:
A -0.133653
B -0.030854
Name: 2, dtype: float64

In [5]: sub_df.iloc[0]['A']
Out[5]: -0.13365288513107493
</code></pre>

blocks|key|187491|text|这些是标量的快速访问|type|unstyled|depth|inlineStyleRanges|entityRanges|data|187492|In+[15]:+df+=+pandas.DataFrame(numpy.random.randn(5,3),columns=list('ABC'))

In+[16]:+df
Out[16]:+
++++++++++A+++++++++B+++++++++C
0+-0.074172+-0.090626++0.038272
1+-0.128545++0.762088+-0.714816
2++0.201498+-0.734963++0.558397
3++1.563307+-1.186415++0.848246
4++0.205171++0.962514++0.037709

In+[17]:+df.iat[0,0]
Out[17]:+-0.074171888537611502

In+[18]:+df.at[0,'A']
Out[18]:+-0.074171888537611502|code-block|syntax|javascript|187493|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|$]]

These are fast access for scalars

<pre><code>In [15]: df = pandas.DataFrame(numpy.random.randn(5,3),columns=list('ABC'))

In [16]: df
Out[16]: 
 A B C
0 -0.074172 -0.090626 0.038272
1 -0.128545 0.762088 -0.714816
2 0.201498 -0.734963 0.558397
3 1.563307 -1.186415 0.848246
4 0.205171 0.962514 0.037709

In [17]: df.iat[0,0]
Out[17]: -0.074171888537611502

In [18]: df.at[0,'A']
Out[18]: -0.074171888537611502
</code></pre>

blocks|key|3018818|text|它看起来像是熊猫10.1/13.1之后的变化|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3018819|在iloc不可用之前，我从10.1升级到了13.1。|3018820|现在，在13.1版本中，iloc[0]['label']获得了单个值数组，而不是标量。|offset|length|style|CODE|3018821|如下所示：|3018822|lastprice=stock.iloc[-1]['Close']|code-block|syntax|javascript|3018823|输出：|3018824|date
2014-02-26+118.2
name:Close,+dtype:+float64|3018825|entityMap^0|0|0|C|G|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|12|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|N|7|14|8|@]|9|@]|A|$O|P]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

It looks like changes after pandas 10.1/13.1

I upgraded from 10.1 to 13.1, before iloc is not available.

Now with 13.1, <code>iloc[0]['label']</code> gets a single value array rather than a scalar.

Like this:

<pre><code>lastprice=stock.iloc[-1]['Close']
</code></pre>

Output:

<pre><code>date
2014-02-26 118.2
name:Close, dtype: float64
</code></pre>

blocks|key|3018827|text|对于iloc不可用的pandas+0.10，过滤DF并获取列VALUE的第一行数据|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3018828|df_filt+=+df[df['C1']+==+C1val+&+df['C2']+==+C2val]
result+=+df_filt.get_value(df_filt.index[0],'VALUE')|code-block|syntax|javascript|3018829|如果筛选的行数超过1行，则获取第一行的值。如果筛选器导致数据框为空，则会出现异常。|3018830|entityMap^0|2|4|O|2|U|5|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

For pandas 0.10, where <code>iloc</code> is unavalable, filter a <code>DF</code> and get the first row data for the column <code>VALUE</code>:

<pre><code>df_filt = df[df['C1'] == C1val &amp; df['C2'] == C2val]
result = df_filt.get_value(df_filt.index[0],'VALUE')
</code></pre>

if there is more then 1 row filtered, obtain the first row value. There will be an exception if the filter result in empty data frame.

blocks|key|187544|text|我不确定这是否是一个好的实践，但是我注意到我也可以通过将该系列转换为float来获得值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|187545|例如：|187546|rate|code-block|syntax|javascript|187547|187548|+3+0.042679|blockquote|187549|187550|名称:+Unemployment_rate，数据类型:+float64|187551|187552|float(rate)|187553|187554|+0.0426789|187555|187556|entityMap^0|Y|5|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|18|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1A|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]|$1|U|3|V|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|W|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|X|3|Y|5|P|7|1E|8|@]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|11|$]]

Not sure if this is a good practice, but I noticed I can also get just the value by casting the series as <code>float</code>.

e.g.

<pre><code>rate
</code></pre>

<blockquote>
 3 0.042679
 
 Name: Unemployment_rate, dtype: float64
</blockquote>

<pre><code>float(rate)
</code></pre>

<blockquote>
 0.0426789
</blockquote>

blocks|key|2806698|text|我找到的最快/最简单的选项如下。501表示行索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2806699|df.at[501,'column_name']
df.get_value(501,'column_name')|code-block|syntax|javascript|2806700|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|$]]

The quickest/easiest options I have found are the following. 501 represents the row index.

<pre><code>df.at[501,'column_name']
df.get_value(501,'column_name')
</code></pre>

blocks|key|186065|text|大多数答案都是使用iloc，这是很好的位置选择。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|186066|如果你需要selection-by-label，loc会更方便。|186067|用于显式获取值的|186068|+(等同于过时的df.get_value(+'A‘，'A’))这也等同于df1.at'a'，‘A’In+55:+df1.loc'a'，‘A’Out55:+0.13200317033032932|blockquote|186069|186070|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/indexing.html#selection-by-label^0|9|4|0|O|3|5|I|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|15|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

Most answers are using <code>iloc</code> which is good for selection by position.

If you need <a href="http://pandas.pydata.org/pandas-docs/stable/indexing.html#selection-by-label" rel="noreferrer">selection-by-label</a> <code>loc</code> would be more convenient.

<blockquote>
 For getting a value explicitly (equiv to deprecated
 df.get_value('a','A'))

<pre><code># this is also equivalent to df1.at['a','A']
In [55]: df1.loc['a', 'A'] 
Out[55]: 0.13200317033032932
</code></pre>
</blockquote>

blocks|key|2806789|text|您可以将1x1数据帧转换为numpy数组，然后访问该数组的第一个也是唯一一个值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2806790|val+=+d2['col_name'].values[0]|code-block|syntax|javascript|2806791|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|$]]

You can turn your 1x1 dataframe into a numpy array, then access the first and only value of that array:

<pre><code>val = d2['col_name'].values[0]
</code></pre>

blocks|key|2806824|text|df_gdp.columns|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2806825|unstyled|2806826|索引(u‘国家’，u‘国家代码’，u‘指标名称’，u‘指标代码’，u'1960'，u'1961'，u'1962'，u'1963'，u'1964'，u'1965'，u'1966'，u'1967'，u'1968'，u'1969'，u'1970'，u'1971'，u'1972'，u'1973'，u‘1974年’，u'1975'，u'1976'，u'1977'，u'1978'，u'1979'，u'1980'，u'1981'，u'1982'，u'1983'，u'1984'，u'1985'，u'1986'，u'1987'，u'1988'，u'1989'，u'1990'，u'1991'，u'1992'，u'1993'，u'1994'，u'1995'，u'1996'，u'1997'，u'1998'，u'1999'，u'2000'，u'2001'，u'2002'，u'2003'，u'2004'，u'2005'，u'2006'，u'2007'，u'2008'，u'2009'，u'2010'，u'2011'，u'2012'，u'2013'，u'2014'，u'2015'，u'2016'，dtype='object')|blockquote|2806827|2806828|2806829|df_gdp[df_gdp["Country+Code"]+==+"USA"]["1996"].values[0]|2806830|2806831|+8100000000000.0|2806832|2806833|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|T|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|U|8|@]|9|@]|A|$]]|$1|I|3|-4|5|E|7|V|8|@]|9|@]|A|$]]|$1|J|3|-4|5|E|7|W|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|E|7|Y|8|@]|9|@]|A|$]]|$1|N|3|O|5|H|7|Z|8|@]|9|@]|A|$]]|$1|P|3|-4|5|E|7|10|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|E|7|11|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>df_gdp.columns
</code></pre>

<blockquote>
 Index([u'Country', u'Country Code', u'Indicator Name', u'Indicator Code',
 u'1960', u'1961', u'1962', u'1963', u'1964', u'1965', u'1966', u'1967',
 u'1968', u'1969', u'1970', u'1971', u'1972', u'1973', u'1974', u'1975',
 u'1976', u'1977', u'1978', u'1979', u'1980', u'1981', u'1982', u'1983',
 u'1984', u'1985', u'1986', u'1987', u'1988', u'1989', u'1990', u'1991',
 u'1992', u'1993', u'1994', u'1995', u'1996', u'1997', u'1998', u'1999',
 u'2000', u'2001', u'2002', u'2003', u'2004', u'2005', u'2006', u'2007',
 u'2008', u'2009', u'2010', u'2011', u'2012', u'2013', u'2014', u'2015',
 u'2016'],
 dtype='object')
</blockquote>

<pre><code>df_gdp[df_gdp["Country Code"] == "USA"]["1996"].values[0]
</code></pre>

<blockquote>
 8100000000000.0
</blockquote>

blocks|key|188108|text|我需要一个单元格的值，通过列名和索引名进行选择。这个解决方案对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|188109|original_conversion_frequency.loc[1,:].values[0]|offset|length|style|CODE|188110|entityMap^0|0|0|1C|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I needed the value of one cell, selected by column and index names.
This solution worked for me:

<code>original_conversion_frequency.loc[1,:].values[0]</code>

blocks|key|2806835|text|它不需要很复杂：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2806836|val+=+df.loc[df.wd==1,+'col_name'].values[0]|code-block|syntax|javascript|2806837|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|$]]

It doesn't need to be complicated:

<pre><code>val = df.loc[df.wd==1, 'col_name'].values[0]
</code></pre>

blocks|key|187630|text|要以JSON+(而不是序列)形式获取整行的值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|187631|row+=+df.iloc[0]|code-block|syntax|javascript|187632|使用如下所示的to_json方法：|offset|length|style|CODE|187633|row.to_json()|187634|entityMap^0|0|0|7|7|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

To get the full row's value as JSON (instead of a Serie):
<pre class="lang-py prettyprint-override"><code>row = df.iloc[0]
</code></pre>
Use the <code>to_json</code> method like bellow:
<pre class="lang-py prettyprint-override"><code>row.to_json()
</code></pre>

blocks|key|2806839|text|将它转换为整数对我来说很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2806840|int(sub_df.iloc[0])|code-block|syntax|javascript|2806841|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|$]]

Converting it to integer worked for me:
<pre><code>int(sub_df.iloc[0])
</code></pre>

blocks|key|3019139|text|我在使用DataFrames和MultiIndexes时遇到过这个问题，我发现squeeze很有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3019140|从文档中：|3019141|3019142|将一维轴对象压缩为标量。|blockquote|3019143|3019144|只有一个元素的Series或DataFrames被压缩成一个标量。单列或单行的DataFrames被压缩成一个系列。否则，对象将保持不变。|3019145|3019146|#+example+for+DataFrame+with+MultiIndex
>+import+pandas+as+pd

>+df+=+pd.DataFrame(
++++++++++++++++++++[
++++++++++++++++++++++++[1,+2,+3],+
++++++++++++++++++++++++[4,+5,+6],+
++++++++++++++++++++++++[7,+8,+9]
++++++++++++++++++++],+
++++++++++++++++++++index=pd.MultiIndex.from_tuples(+[('i',+1),+('ii',+2),+('iii',+3)]+),
++++++++++++++++++++columns=pd.MultiIndex.from_tuples(+[('A',+'a'),+('B',+'b'),+('C',+'c')]+)
)

>+df
+++++++A++B++C
+++++++a++b++c
i+++1++1++2++3
ii++2++4++5++6
iii+3++7++8++9

>+df.loc['ii',+'B']
+++b
2++5

>+df.loc['ii',+'B'].squeeze()
5|code-block|syntax|javascript|3019147|请注意，虽然df.at[]也可以工作(如果您不需要使用条件)，但是您仍然需要指定MultiIndex的所有级别。|style|CODE|3019148|示例：|3019149|>+df.at[('ii',+2),+('B',+'b')]
5|3019150|我有一个具有6级索引和2级列的DataFrame，因此只需指定外部级别是非常有帮助的。|3019151|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.squeeze.html^0|13|7|0|0|0|0|0|0|0|0|0|6|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|D|3|E|5|6|7|1E|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|1F|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|1G|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|1H|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|1I|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|1J|8|@]|9|@]|C|$]]|$1|N|3|O|5|P|7|1K|8|@]|9|@]|C|$Q|R]]|$1|S|3|T|5|6|7|1L|8|@$A|1M|B|1N|U|V]]|9|@]|C|$]]|$1|W|3|X|5|6|7|1O|8|@]|9|@]|C|$]]|$1|Y|3|Z|5|P|7|1P|8|@]|9|@]|C|$Q|R]]|$1|10|3|11|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|12|3|-4|5|6|7|1R|8|@]|9|@]|C|$]]]|13|$14|$5|15|16|17|C|$18|19]]]]

I've run across this when using DataFrames with MultiIndexes and found <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.squeeze.html" rel="nofollow noreferrer">squeeze</a> useful.
From the docs:
<blockquote>
Squeeze 1 dimensional axis objects into scalars.
Series or DataFrames with a single element are squeezed to a scalar.
DataFrames with a single column or a single row are squeezed to a
Series. Otherwise the object is unchanged.
</blockquote>
<pre><code># example for DataFrame with MultiIndex
&gt; import pandas as pd

&gt; df = pd.DataFrame(
 [
 [1, 2, 3], 
 [4, 5, 6], 
 [7, 8, 9]
 ], 
 index=pd.MultiIndex.from_tuples( [('i', 1), ('ii', 2), ('iii', 3)] ),
 columns=pd.MultiIndex.from_tuples( [('A', 'a'), ('B', 'b'), ('C', 'c')] )
)

&gt; df
 A B C
 a b c
i 1 1 2 3
ii 2 4 5 6
iii 3 7 8 9

&gt; df.loc['ii', 'B']
 b
2 5

&gt; df.loc['ii', 'B'].squeeze()
5
</code></pre>
Note that while <code>df.at[]</code> also works (if you aren't needing to use conditionals) you then still AFAIK need to specify all levels of the MultiIndex.
Example:
<pre><code>&gt; df.at[('ii', 2), ('B', 'b')]
5
</code></pre>
I have a DataFrame with a 6-level index and 2-level columns, so only having to specify the outer level is quite helpful.

blocks|key|188156|text|这已经很旧了，但是现在你可以通过简单的如下操作来修复它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|188157|val+=+float(d2['col_name'].iloc[0])|code-block|syntax|javascript|188158|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|$]]

This is quite old by now but as of today you can fix it by simply doing
<pre><code>val = float(d2['col_name'].iloc[0])
</code></pre>

I have constructed a condition that extract exactly one row from my data frame:

<pre><code>d2 = df[(df['l_ext']==l_ext) &amp; (df['item']==item) &amp; (df['wn']==wn) &amp; (df['wd']==1)]
</code></pre>

Now I would like to take a value from a particular column:

<pre><code>val = d2['col_name']
</code></pre>

But as a result I get a data frame that contains one row and one column (i.e. one cell). It is not what I need. I need one value (one float number). How can I do it in pandas?

How to get a value from a cell of a dataframe?

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我构造了一个条件，从我的数据框中恰好提取一行：d2 = df[(df['l_ext']==l_ext) & (df['item']==item) & (df['wn']==wn) & (df['wd']==1)]现在我想从一个特定的列中获取一个值：val = d2['col_name']但结果是我得到了一个包含一行和一...

问如何从数据帧的单元格中获取值？
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