将@piRSquared的答案扩展到问题10的另一个版本

问题10.1

DataFrame:

d = data = {'A': {0: 1, 1: 1, 2: 1, 3: 2, 4: 2, 5: 3, 6: 5},
 'B': {0: 'a', 1: 'b', 2: 'c', 3: 'a', 4: 'b', 5: 'a', 6: 'c'}}
df = pd.DataFrame(d)

   A  B
0  1  a
1  1  b
2  1  c
3  2  a
4  2  b
5  3  a
6  5  c

输出:

   0     1     2
A
1  a     b     c
2  a     b  None
3  a  None  None
5  c  None  None

---

使用df。groupby和pd.Series.tolist

t = df.groupby('A')['B'].apply(list)
out = pd.DataFrame(t.tolist(),index=t.index)
out
   0     1     2
A
1  a     b     c
2  a     b  None
3  a  None  None
5  c  None  None

或 使用pd是更好的选择。使用df.squeeze的数据透视表。

t = df.pivot_table(index='A',values='B',aggfunc=list).squeeze()
out = pd.DataFrame(t.tolist(),index=t.index)

调用reset_index()(连同add_suffix())

通常情况下，reset_index()在调用数据透视表或pivot之后是需要的。例如，要进行以下转换(其中一列变成列名)

你使用下面的代码，在pivot之后，你为新创建的列名添加前缀，并将索引(在本例中是“movies”)转换回一个列，并删除轴名的名称:

df.pivot(*df).add_prefix('week_').reset_index().rename_axis(columns=None)

---

正如其他答案所提到的，“pivot”可能指的是两种不同的操作:

未堆叠聚合(即使groupby的结果。gg。) 重塑(类似于Excel中的pivot，在numpy中重塑或在R中pivot_wider)

1. 聚合

数据透视表或交叉表只是groupby的未堆叠结果。gg操作。事实上，源代码显示，在引擎盖下，以下是正确的:

数据透视表= groupby + unstack(阅读这里了解更多信息) 交叉表=数据透视表

注意:您可以使用列名称列表作为索引，列和值参数。

df.groupby(rows+cols)[vals].agg(aggfuncs).unstack(cols)
# equivalently,
df.pivot_table(vals, rows, cols, aggfuncs)

1.1. Crosstab是pivot_table的特殊情况;因此，groupby + unstack

以下是等价的:

pd。crosstab (df[“可乐”,df[‘colB]) df。标签=“可乐”、标签=“科尔”、aggfunc=“尺寸”、文件价值=0) df。groupby([可乐’,‘colB)大小()。unstack (fill_value = 0)

注意pd。Crosstab的开销要大得多，所以它比pivot_table和groupby + unstack慢得多。事实上，正如这里提到的，pivot_table也比groupby + unstack慢。

2. 重塑

Pivot是pivot_table的一个更有限的版本，它的目的是将一个长数据帧重新塑造成一个长数据帧。

df.set_index(rows+cols)[vals].unstack(cols)
# equivalently, 
df.pivot(rows, cols, vals)

2.1. 如问题10所示增加行/列

你也可以将问题10中的见解应用到多列枢轴运算中。有两种情况:

"long-to-long": reshape by augmenting the indices Code: df = pd.DataFrame({'A': [1, 1, 1, 2, 2, 2], 'B': [*'xxyyzz'], 'C': [*'CCDCDD'], 'E': [100, 200, 300, 400, 500, 600]}) rows, cols, vals = ['A', 'B'], ['C'], 'E' # using pivot syntax df1 = ( df.assign(ix=df.groupby(rows+cols).cumcount()) .pivot([*rows, 'ix'], cols, vals) .fillna(0, downcast='infer') .droplevel(-1).reset_index().rename_axis(columns=None) ) # equivalently, using set_index + unstack syntax df1 = ( df .set_index([*rows, df.groupby(rows+cols).cumcount(), *cols])[vals] .unstack(fill_value=0) .droplevel(-1).reset_index().rename_axis(columns=None) ) "long-to-wide": reshape by augmenting the columns Code: df1 = ( df.assign(ix=df.groupby(rows+cols).cumcount()) .pivot(rows, [*cols, 'ix'])[vals] .fillna(0, downcast='infer') ) df1 = df1.set_axis([f"{c[0]}_{c[1]}" for c in df1], axis=1).reset_index() # equivalently, using the set_index + unstack syntax df1 = ( df .set_index([*rows, df.groupby(rows+cols).cumcount(), *cols])[vals] .unstack([-1, *range(-2, -len(cols)-2, -1)], fill_value=0) ) df1 = df1.set_axis([f"{c[0]}_{c[1]}" for c in df1], axis=1).reset_index() minimum case using the set_index + unstack syntax: Code: df1 = df.set_index(['A', df.groupby('A').cumcount()])['E'].unstack(fill_value=0).add_prefix('Col').reset_index()

---

1 pivot_table() aggregates the values and unstacks it. Specifically, it creates a single flat list out of index and columns, calls groupby() with this list as the grouper and aggregates using the passed aggregator methods (the default is mean). Then after aggregation, it calls unstack() by the list of columns. So internally, pivot_table = groupby + unstack. Moreover, if fill_value is passed, fillna() is called. In other words, the method that produces pv_1 is the same as the method that produces gb_1 in the example below. pv_1 = df.pivot_table(index=rows, columns=cols, values=vals, aggfunc=aggfuncs, fill_value=0) # internal operation of `pivot_table()` gb_1 = df.groupby(rows+cols)[vals].agg(aggfuncs).unstack(cols).fillna(0, downcast="infer") pv_1.equals(gb_1) # True 2 crosstab() calls pivot_table(), i.e., crosstab = pivot_table. Specifically, it builds a DataFrame out of the passed arrays of values, filters it by the common indices and calls pivot_table(). It's more limited than pivot_table() because it only allows a one-dimensional array-like as values, unlike pivot_table() that can have multiple columns as values.

为了更好地理解函数枢轴是如何工作的，您可以查看Pandas文档中的示例。然而，如果你有重复的索引-列(foo-bar)组合(就像第二个例子中的df)， pivot将失败:

与枢轴相反，函数pivot_table默认支持使用平均值函数进行数据聚合。下面是一个求和聚合函数的例子:

pandas中的pivot函数与excel中的pivot操作具有相同的功能。我们可以将数据集从长格式转换为宽格式。

我们来举个例子

我们希望将数据集转换为一种形式，使每个国家成为一列，新确诊病例作为对应国家的值。我们可以使用枢轴函数来执行这个数据操作。

Pivot数据集

pivot_df = pd.pivot(df, index =['Date'], columns ='Country', values =['NewConfirmed'])
## renaming the columns  
pivot_df.columns = df['Country'].sort_values().unique()

我们可以通过重置索引将新列放到与索引列Data相同的级别。

重置索引以修改列级别

Pivot_df = Pivot_df .reset_index()

将@piRSquared的答案扩展到问题10的另一个版本

问题10.1

DataFrame:

d = data = {'A': {0: 1, 1: 1, 2: 1, 3: 2, 4: 2, 5: 3, 6: 5},
 'B': {0: 'a', 1: 'b', 2: 'c', 3: 'a', 4: 'b', 5: 'a', 6: 'c'}}
df = pd.DataFrame(d)

   A  B
0  1  a
1  1  b
2  1  c
3  2  a
4  2  b
5  3  a
6  5  c

输出:

   0     1     2
A
1  a     b     c
2  a     b  None
3  a  None  None
5  c  None  None

---

使用df。groupby和pd.Series.tolist

t = df.groupby('A')['B'].apply(list)
out = pd.DataFrame(t.tolist(),index=t.index)
out
   0     1     2
A
1  a     b     c
2  a     b  None
3  a  None  None
5  c  None  None

或 使用pd是更好的选择。使用df.squeeze的数据透视表。

t = df.pivot_table(index='A',values='B',aggfunc=list).squeeze()
out = pd.DataFrame(t.tolist(),index=t.index)

下面是我们可以用来支点的习语列表

pd.DataFrame.pivot_table A glorified version of groupby with more intuitive API. For many people, this is the preferred approach. And it is the intended approach by the developers. Specify row level, column levels, values to be aggregated, and function(s) to perform aggregations. pd.DataFrame.groupby + pd.DataFrame.unstack Good general approach for doing just about any type of pivot You specify all columns that will constitute the pivoted row levels and column levels in one group by. You follow that by selecting the remaining columns you want to aggregate and the function(s) you want to perform the aggregation. Finally, you unstack the levels that you want to be in the column index. pd.DataFrame.set_index + pd.DataFrame.unstack Convenient and intuitive for some (myself included). Cannot handle duplicate grouped keys. Similar to the groupby paradigm, we specify all columns that will eventually be either row or column levels and set those to be the index. We then unstack the levels we want in the columns. If either the remaining index levels or column levels are not unique, this method will fail. pd.DataFrame.pivot Very similar to set_index in that it shares the duplicate key limitation. The API is very limited as well. It only takes scalar values for index, columns, values. Similar to the pivot_table method in that we select rows, columns, and values on which to pivot. However, we cannot aggregate and if either rows or columns are not unique, this method will fail. pd.crosstab This a specialized version of pivot_table and in its purest form is the most intuitive way to perform several tasks. pd.factorize + np.bincount This is a highly advanced technique that is very obscure but is very fast. It cannot be used in all circumstances, but when it can be used and you are comfortable using it, you will reap the performance rewards. pd.get_dummies + pd.DataFrame.dot I use this for cleverly performing cross tabulation.

参见:

重塑和透视表- pandas用户指南

---

问题1

为什么我得到ValueError:索引包含重复的条目，不能重塑

出现这种情况是因为pandas试图重新索引具有重复条目的列或索引对象。可以使用不同的方法来执行枢轴旋转。当要求它旋转的键有重复时，其中一些不太适合。例如:考虑pd.DataFrame.pivot。我知道有重复的条目共享row和col值:

df.duplicated(['row', 'col']).any()

True

所以当我使用主元

df.pivot(index='row', columns='col', values='val0')

我得到了上面提到的错误。事实上，当我尝试执行相同的任务时，我得到了相同的错误:

df.set_index(['row', 'col'])['val0'].unstack()

---

例子

我接下来要做的是用pd。dataframe。pivot_table来回答每个问题。然后，我将提供执行相同任务的替代方案。

问题2和3

我如何枢轴df，使col值是列，行值是索引，val0的平均值是值?

pd.DataFrame.pivot_table df.pivot_table ( Values ='val0'， index='row'， columns='col'， aggfunc =“的意思是”) colcol0 col1 col2 col3 col4 行 row0 0.77 0.605 NaN 0.860 0.65 第二行0.13 NaN 0.395 0.500 0.25 第三行NaN 0.310 NaN 0.545 NaN row4 NaN 0.100 0.395 0.760 0.24 aggfunc='mean'是默认值，我不需要设置它。我把它写出来是为了更明确。

我怎样才能使缺失值为0呢?

pd.DataFrame.pivot_table fill_value is not set by default. I tend to set it appropriately. In this case I set it to 0. df.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc='mean') col col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 row2 0.13 0.000 0.395 0.500 0.25 row3 0.00 0.310 0.000 0.545 0.00 row4 0.00 0.100 0.395 0.760 0.24 pd.DataFrame.groupby df.groupby(['row', 'col'])['val0'].mean().unstack(fill_value=0) pd.crosstab pd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc='mean').fillna(0)

---

问题4

我能说点别的吗，比如求和?

pd.DataFrame.pivot_table df.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc='sum') col col0 col1 col2 col3 col4 row row0 0.77 1.21 0.00 0.86 0.65 row2 0.13 0.00 0.79 0.50 0.50 row3 0.00 0.31 0.00 1.09 0.00 row4 0.00 0.10 0.79 1.52 0.24 pd.DataFrame.groupby df.groupby(['row', 'col'])['val0'].sum().unstack(fill_value=0) pd.crosstab pd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc='sum').fillna(0)

---

问题5

我能一次做更多的聚合吗?

注意，对于数据透视表和交叉表，我需要传递可调用对象的列表。另一方面，groupby。Agg能够为有限数量的特殊函数接受字符串。groupby。Agg也会使用我们传递给其他函数的相同的可调用对象，但利用字符串函数名通常更有效，因为可以获得效率。

pd.DataFrame.pivot_table df.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc=[np.size, np.mean]) size mean col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 1 2 0 1 1 0.77 0.605 0.000 0.860 0.65 row2 1 0 2 1 2 0.13 0.000 0.395 0.500 0.25 row3 0 1 0 2 0 0.00 0.310 0.000 0.545 0.00 row4 0 1 2 2 1 0.00 0.100 0.395 0.760 0.24 pd.DataFrame.groupby df.groupby(['row', 'col'])['val0'].agg(['size', 'mean']).unstack(fill_value=0) pd.crosstab pd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc=[np.size, np.mean]).fillna(0, downcast='infer')

---

问题6

我可以聚合多个值列吗?

pd.DataFrame.pivot_table we pass values=['val0', 'val1'] but we could've left that off completely df.pivot_table( values=['val0', 'val1'], index='row', columns='col', fill_value=0, aggfunc='mean') val0 val1 col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 0.01 0.745 0.00 0.010 0.02 row2 0.13 0.000 0.395 0.500 0.25 0.45 0.000 0.34 0.440 0.79 row3 0.00 0.310 0.000 0.545 0.00 0.00 0.230 0.00 0.075 0.00 row4 0.00 0.100 0.395 0.760 0.24 0.00 0.070 0.42 0.300 0.46 pd.DataFrame.groupby df.groupby(['row', 'col'])['val0', 'val1'].mean().unstack(fill_value=0)

---

问题7

我可以用多列细分吗?

pd.DataFrame.pivot_table df.pivot_table( values='val0', index='row', columns=['item', 'col'], fill_value=0, aggfunc='mean') item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 row row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.605 0.86 0.65 row2 0.35 0.00 0.37 0.00 0.00 0.44 0.00 0.00 0.13 0.000 0.50 0.13 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.000 0.28 0.00 row4 0.15 0.64 0.00 0.00 0.10 0.64 0.88 0.24 0.00 0.000 0.00 0.00 pd.DataFrame.groupby df.groupby( ['row', 'item', 'col'] )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1)

---

问题8

我可以用多列细分吗?

pd.DataFrame.pivot_table df.pivot_table( values='val0', index=['key', 'row'], columns=['item', 'col'], fill_value=0, aggfunc='mean') item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 key row key0 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.86 0.00 row2 0.00 0.00 0.37 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.50 0.00 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.00 0.00 0.00 row4 0.15 0.64 0.00 0.00 0.00 0.00 0.00 0.24 0.00 0.00 0.00 0.00 key1 row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.81 0.00 0.65 row2 0.35 0.00 0.00 0.00 0.00 0.44 0.00 0.00 0.00 0.00 0.00 0.13 row3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.28 0.00 row4 0.00 0.00 0.00 0.00 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 key2 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.40 0.00 0.00 row2 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.13 0.00 0.00 0.00 row4 0.00 0.00 0.00 0.00 0.00 0.64 0.88 0.00 0.00 0.00 0.00 0.00 pd.DataFrame.groupby df.groupby( ['key', 'row', 'item', 'col'] )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1) pd.DataFrame.set_index because the set of keys are unique for both rows and columns df.set_index( ['key', 'row', 'item', 'col'] )['val0'].unstack(['item', 'col']).fillna(0).sort_index(1)

---

问题9

我可以聚合列和行一起出现的频率，也就是“交叉制表”吗?

pd.DataFrame.pivot_table df.pivot_table(index='row', columns='col', fill_value=0, aggfunc='size') col col0 col1 col2 col3 col4 row row0 1 2 0 1 1 row2 1 0 2 1 2 row3 0 1 0 2 0 row4 0 1 2 2 1 pd.DataFrame.groupby df.groupby(['row', 'col'])['val0'].size().unstack(fill_value=0) pd.crosstab pd.crosstab(df['row'], df['col']) pd.factorize + np.bincount # get integer factorization `i` and unique values `r` # for column `'row'` i, r = pd.factorize(df['row'].values) # get integer factorization `j` and unique values `c` # for column `'col'` j, c = pd.factorize(df['col'].values) # `n` will be the number of rows # `m` will be the number of columns n, m = r.size, c.size # `i * m + j` is a clever way of counting the # factorization bins assuming a flat array of length # `n * m`. Which is why we subsequently reshape as `(n, m)` b = np.bincount(i * m + j, minlength=n * m).reshape(n, m) # BTW, whenever I read this, I think 'Bean, Rice, and Cheese' pd.DataFrame(b, r, c) col3 col2 col0 col1 col4 row3 2 0 0 1 0 row2 1 2 1 0 2 row0 1 0 1 2 1 row4 2 2 0 1 1 pd.get_dummies pd.get_dummies(df['row']).T.dot(pd.get_dummies(df['col'])) col0 col1 col2 col3 col4 row0 1 2 0 1 1 row2 1 0 2 1 2 row3 0 1 0 2 0 row4 0 1 2 2 1

---

问题10

我如何转换一个数据帧从长到宽的枢轴上只有两个 列?

DataFrame.pivot The first step is to assign a number to each row - this number will be the row index of that value in the pivoted result. This is done using GroupBy.cumcount: df2.insert(0, 'count', df2.groupby('A').cumcount()) df2 count A B 0 0 a 0 1 1 a 11 2 2 a 2 3 3 a 11 4 0 b 10 5 1 b 10 6 2 b 14 7 0 c 7 The second step is to use the newly created column as the index to call DataFrame.pivot. df2.pivot(*df2) # df2.pivot(index='count', columns='A', values='B') A a b c count 0 0.0 10.0 7.0 1 11.0 10.0 NaN 2 2.0 14.0 NaN 3 11.0 NaN NaN DataFrame.pivot_table Whereas DataFrame.pivot only accepts columns, DataFrame.pivot_table also accepts arrays, so the GroupBy.cumcount can be passed directly as the index without creating an explicit column. df2.pivot_table(index=df2.groupby('A').cumcount(), columns='A', values='B') A a b c 0 0.0 10.0 7.0 1 11.0 10.0 NaN 2 2.0 14.0 NaN 3 11.0 NaN NaN

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问题11

我如何在主轴后将多个索引平展为单个索引

如果列类型对象与字符串连接

df.columns = df.columns.map('|'.join)

其他的格式

df.columns = df.columns.map('{0[0]}|{0[1]}'.format)