熊猫-获取给定列的第一行值

问题:熊猫-获取给定列的第一行值

这似乎是一个非常简单的问题……但是我没有看到我期望的简单答案。

那么,如何获得Pandas中给定列的第n行的值?(我对第一行特别感兴趣,但也对更通用的做法感兴趣)。

例如,假设我想将Btime中的1.2值作为变量。

什么是正确的方法?

df_test =

  ATime   X   Y   Z   Btime  C   D   E
0    1.2  2  15   2    1.2  12  25  12
1    1.4  3  12   1    1.3  13  22  11
2    1.5  1  10   6    1.4  11  20  16
3    1.6  2   9  10    1.7  12  29  12
4    1.9  1   1   9    1.9  11  21  19
5    2.0  0   0   0    2.0   8  10  11
6    2.4  0   0   0    2.4  10  12  15

This seems like a ridiculously easy question… but I’m not seeing the easy answer I was expecting.

So, how do I get the value at an nth row of a given column in Pandas? (I am particularly interested in the first row, but would be interested in a more general practice as well).

For example, let’s say I want to pull the 1.2 value in Btime as a variable.

Whats the right way to do this?

df_test =

  ATime   X   Y   Z   Btime  C   D   E
0    1.2  2  15   2    1.2  12  25  12
1    1.4  3  12   1    1.3  13  22  11
2    1.5  1  10   6    1.4  11  20  16
3    1.6  2   9  10    1.7  12  29  12
4    1.9  1   1   9    1.9  11  21  19
5    2.0  0   0   0    2.0   8  10  11
6    2.4  0   0   0    2.4  10  12  15

回答 0

要选择该ith行,请使用iloc

In [31]: df_test.iloc[0]
Out[31]: 
ATime     1.2
X         2.0
Y        15.0
Z         2.0
Btime     1.2
C        12.0
D        25.0
E        12.0
Name: 0, dtype: float64

要在Btime列中选择第i个值,可以使用:

In [30]: df_test['Btime'].iloc[0]
Out[30]: 1.2

df_test['Btime'].iloc[0](推荐)和之间有区别df_test.iloc[0]['Btime']

DataFrames将数据存储在基于列的块中(每个块具有一个dtype)。如果先按列选择,则可以返回视图(比返回副本要快),并且保留原始dtype。相反,如果首先选择按行,并且DataFrame的列具有不同的dtype,则Pandas 将数据复制到新的Object dtype 系列中。因此,选择列比选择行要快一些。因此,虽然 df_test.iloc[0]['Btime']作品,df_test['Btime'].iloc[0]是多一点点效率。

在分配方面,两者之间存在很大差异。 df_test['Btime'].iloc[0] = x影响df_test,但df_test.iloc[0]['Btime'] 可能不会。有关原因的说明,请参见下文。由于索引顺序的细微差别会在行为上产生很大差异,因此最好使用单个索引分配:

df.iloc[0, df.columns.get_loc('Btime')] = x

df.iloc[0, df.columns.get_loc('Btime')] = x (推荐的):

为DataFrame分配新值的推荐方法避免链接索引,而应使用andrew所示的方法,

df.loc[df.index[n], 'Btime'] = x

要么

df.iloc[n, df.columns.get_loc('Btime')] = x

后一种方法要快一些,因为df.loc必须将行和列标签转换为位置索引,因此,如果使用df.iloc替代方法,则转换的必要性要少一些 。


df['Btime'].iloc[0] = x 可行,但不建议:

尽管这可行,但是它利用了当前实现DataFrames的方式。不能保证熊猫将来会以这种方式工作。特别是,它利用了以下事实:(当前)df['Btime']始终返回视图(而不是副本),因此df['Btime'].iloc[n] = x可用于在的列的第n个位置分配新值。Btimedf

由于Pandas无法明确保证索引器何时返回视图还是副本,因此使用链式索引的赋值通常会引发,SettingWithCopyWarning即使在这种情况下,赋值可以成功修改df

In [22]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [24]: df['bar'] = 100
In [25]: df['bar'].iloc[0] = 99
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

In [26]: df
Out[26]: 
  foo  bar
0   A   99  <-- assignment succeeded
2   B  100
1   C  100

df.iloc[0]['Btime'] = x 不起作用:

相比之下,with的分配df.iloc[0]['bar'] = 123不起作用,因为df.iloc[0]正在返回副本:

In [66]: df.iloc[0]['bar'] = 123
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

In [67]: df
Out[67]: 
  foo  bar
0   A   99  <-- assignment failed
2   B  100
1   C  100

警告:我之前曾建议过df_test.ix[i, 'Btime']。但这不能保证为您提供ith值,因为在尝试按位置索引之前先尝试ix标签索引。因此,如果DataFrame的整数索引不是从0开始的排序顺序,则using 将返回标有标签的行,而不是该行。例如,ix[i] iith

In [1]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])

In [2]: df
Out[2]: 
  foo
0   A
2   B
1   C

In [4]: df.ix[1, 'foo']
Out[4]: 'C'

To select the ith row, use iloc:

In [31]: df_test.iloc[0]
Out[31]: 
ATime     1.2
X         2.0
Y        15.0
Z         2.0
Btime     1.2
C        12.0
D        25.0
E        12.0
Name: 0, dtype: float64

To select the ith value in the Btime column you could use:

In [30]: df_test['Btime'].iloc[0]
Out[30]: 1.2

There is a difference between df_test['Btime'].iloc[0] (recommended) and df_test.iloc[0]['Btime']:

DataFrames store data in column-based blocks (where each block has a single dtype). If you select by column first, a view can be returned (which is quicker than returning a copy) and the original dtype is preserved. In contrast, if you select by row first, and if the DataFrame has columns of different dtypes, then Pandas copies the data into a new Series of object dtype. So selecting columns is a bit faster than selecting rows. Thus, although df_test.iloc[0]['Btime'] works, df_test['Btime'].iloc[0] is a little bit more efficient.

There is a big difference between the two when it comes to assignment. df_test['Btime'].iloc[0] = x affects df_test, but df_test.iloc[0]['Btime'] may not. See below for an explanation of why. Because a subtle difference in the order of indexing makes a big difference in behavior, it is better to use single indexing assignment:

df.iloc[0, df.columns.get_loc('Btime')] = x

df.iloc[0, df.columns.get_loc('Btime')] = x (recommended):

The recommended way to assign new values to a DataFrame is to avoid chained indexing, and instead use the method shown by andrew,

df.loc[df.index[n], 'Btime'] = x

or

df.iloc[n, df.columns.get_loc('Btime')] = x

The latter method is a bit faster, because df.loc has to convert the row and column labels to positional indices, so there is a little less conversion necessary if you use df.iloc instead.


df['Btime'].iloc[0] = x works, but is not recommended:

Although this works, it is taking advantage of the way DataFrames are currently implemented. There is no guarantee that Pandas has to work this way in the future. In particular, it is taking advantage of the fact that (currently) df['Btime'] always returns a view (not a copy) so df['Btime'].iloc[n] = x can be used to assign a new value at the nth location of the Btime column of df.

Since Pandas makes no explicit guarantees about when indexers return a view versus a copy, assignments that use chained indexing generally always raise a SettingWithCopyWarning even though in this case the assignment succeeds in modifying df:

In [22]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [24]: df['bar'] = 100
In [25]: df['bar'].iloc[0] = 99
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

In [26]: df
Out[26]: 
  foo  bar
0   A   99  <-- assignment succeeded
2   B  100
1   C  100

df.iloc[0]['Btime'] = x does not work:

In contrast, assignment with df.iloc[0]['bar'] = 123 does not work because df.iloc[0] is returning a copy:

In [66]: df.iloc[0]['bar'] = 123
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

In [67]: df
Out[67]: 
  foo  bar
0   A   99  <-- assignment failed
2   B  100
1   C  100

Warning: I had previously suggested df_test.ix[i, 'Btime']. But this is not guaranteed to give you the ith value since ix tries to index by label before trying to index by position. So if the DataFrame has an integer index which is not in sorted order starting at 0, then using ix[i] will return the row labeled i rather than the ith row. For example,

In [1]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])

In [2]: df
Out[2]: 
  foo
0   A
2   B
1   C

In [4]: df.ix[1, 'foo']
Out[4]: 'C'

回答 1

请注意,@ unutbu的答案是正确的,直到您想将值设置为新值,否则如果您的数据框是视图,则该答案将不起作用。

In [4]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [5]: df['bar'] = 100
In [6]: df['bar'].iloc[0] = 99
/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/pandas-0.16.0_19_g8d2818e-py2.7-macosx-10.9-x86_64.egg/pandas/core/indexing.py:118: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

See the the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

可以同时在设置和获取上使用的另一种方法是:

In [7]: df.loc[df.index[0], 'foo']
Out[7]: 'A'
In [8]: df.loc[df.index[0], 'bar'] = 99
In [9]: df
Out[9]:
  foo  bar
0   A   99
2   B  100
1   C  100

Note that the answer from @unutbu will be correct until you want to set the value to something new, then it will not work if your dataframe is a view.

In [4]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [5]: df['bar'] = 100
In [6]: df['bar'].iloc[0] = 99
/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/pandas-0.16.0_19_g8d2818e-py2.7-macosx-10.9-x86_64.egg/pandas/core/indexing.py:118: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

See the the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

Another approach that will consistently work with both setting and getting is:

In [7]: df.loc[df.index[0], 'foo']
Out[7]: 'A'
In [8]: df.loc[df.index[0], 'bar'] = 99
In [9]: df
Out[9]:
  foo  bar
0   A   99
2   B  100
1   C  100

回答 2

另一种方法是:

first_value = df['Btime'].values[0]

这种方式似乎比使用更快.iloc

In [1]: %timeit -n 1000 df['Btime'].values[20]
5.82 µs ± 142 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [2]: %timeit -n 1000 df['Btime'].iloc[20]
29.2 µs ± 1.28 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Another way to do this:

first_value = df['Btime'].values[0]

This way seems to be faster than using .iloc:

In [1]: %timeit -n 1000 df['Btime'].values[20]
5.82 µs ± 142 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [2]: %timeit -n 1000 df['Btime'].iloc[20]
29.2 µs ± 1.28 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

回答 3

  1. df.iloc[0].head(1) -仅从整个第一行开始的第一个数据集。
  2. df.iloc[0] -整个列的第一行。
  1. df.iloc[0].head(1) – First data set only from entire first row.
  2. df.iloc[0] – Entire First row in column.

回答 4

通常,如果您想从J列中获取前N行,最好的方法是:pandas dataframe

data = dataframe[0:N][:,J]

In a general way, if you want to pick up the first N rows from the J column from pandas dataframe the best way to do this is:

data = dataframe[0:N][:,J]

回答 5

为了从列“ test”和第1行获取例如值,它的工作原理如下

df[['test']].values[0][0]

因为只df[['test']].values[0]给一个数组

To get e.g the value from column ‘test’ and row 1 it works like

df[['test']].values[0][0]

as only df[['test']].values[0] gives back a array


回答 6

获取第一行并保留索引的另一种方法:

x = df.first('d') # Returns the first day. '3d' gives first three days.

Another way of getting the first row and preserving the index:

x = df.first('d') # Returns the first day. '3d' gives first three days.