Question

也许groupby是错误的做法。似乎应该可以工作，但我没有看到它......

我想根据结果对事件进行分组。这是我的DataFrame（df）：

Event   SUCCESS FAILED
Run     2       1
Walk    0       1

这是我想要的结果：

grouped = df['Status'].groupby(df['Event'])

我正在尝试制作一个分组对象，但我无法弄清楚如何调用它来显示我想要的内容。

var Person = mongoose.model('Person', yourSchema);

// find each person with a last name matching 'Ghost', selecting the `name` and `occupation` fields
Person.findOne({ 'name.last': 'Ghost' }, 'name occupation', function (err, person) {
  if (err) return handleError(err);
  console.log('%s %s is a %s.', person.name.first, person.name.last, person.occupation) // Space Ghost is a talk show host.
})

Answer 1

试试这个：

 pd.crosstab(df.Event, df.Status)

Status  FAILED  SUCCESS
Event                  
Run          0        2
Walk         1        1


len("df.groupby('Event').Status.value_counts().unstack().fillna(0)")
61

len("df.pivot_table(index='Event', columns='Status', aggfunc=len, fill_value=0)")
74

len("pd.crosstab(df.Event, df.Status)")
32

Answer 2

另一种解决方案，使用pivot_table()方法：

In [5]: df.pivot_table(index='Event', columns='Status', aggfunc=len, fill_value=0)
Out[5]:
Status  FAILED  SUCCESS
Event
Run          0        2
Walk         1        1

针对700K DF的时间：

In [74]: df.shape
Out[74]: (700000, 2)

In [75]: # (c) Merlin

In [76]: %%timeit
   ....: pd.crosstab(df.Event, df.Status)
   ....:
1 loop, best of 3: 333 ms per loop

In [77]: # (c) piRSquared

In [78]: %%timeit
   ....: df.groupby('Event').Status.value_counts().unstack().fillna(0)
   ....:
1 loop, best of 3: 325 ms per loop

In [79]: # (c) MaxU

In [80]: %%timeit
   ....: df.pivot_table(index='Event', columns='Status',
   ....:                aggfunc=len, fill_value=0)
   ....:
1 loop, best of 3: 367 ms per loop

In [81]: # (c) ayhan

In [82]: %%timeit
   ....: (df.assign(ones = np.ones(len(df)))
   ....:    .pivot_table(index='Event', columns='Status',
   ....:                 aggfunc=np.sum, values = 'ones')
   ....: )
   ....:
1 loop, best of 3: 264 ms per loop

In [83]: # (c) Divakar

In [84]: %%timeit
   ....: unq1,ID1 = np.unique(df['Event'],return_inverse=True)
   ....: unq2,ID2 = np.unique(df['Status'],return_inverse=True)
   ....: # Get linear indices/tags corresponding to grouped headers
   ....: tag = ID1*(ID2.max()+1) + ID2
   ....: # Setup 2D Numpy array equivalent of expected Dataframe
   ....: out = np.zeros((len(unq1),len(unq2)),dtype=int)
   ....: unqID, count = np.unique(tag,return_counts=True)
   ....: np.put(out,unqID,count)
   ....: # Finally convert to Dataframe
   ....: df_out = pd.DataFrame(out,columns=unq2)
   ....: df_out.index = unq1
   ....:
1 loop, best of 3: 2.25 s per loop

结论：@ ayhan的解决方案目前胜出：

(df.assign(ones = np.ones(len(df)))
   .pivot_table(index='Event', columns='Status', values = 'ones',
                aggfunc=np.sum, fill_value=0)
)

Answer 3

我会这样做：

df.groupby('Event').Status.value_counts().unstack().fillna(0)

时序

Answer 4

这是一种基于NumPy的方法 -

# Get unique header strings for input dataframes
unq1,ID1 = np.unique(df['Event'],return_inverse=True)
unq2,ID2 = np.unique(df['Status'],return_inverse=True)

# Get linear indices/tags corresponding to grouped headers
tag = ID1*(ID2.max()+1) + ID2

# Setup 2D Numpy array equivalent of expected Dataframe
out = np.zeros((len(unq1),len(unq2)),dtype=int)
unqID, count = np.unique(tag,return_counts=True)
np.put(out,unqID,count)

# Finally convert to Dataframe
df_out = pd.DataFrame(out,columns=unq2)
df_out.index = unq1

示例输入，输出更通用的情况 -

In [179]: df
Out[179]: 
  Event   Status
0   Sit     PASS
1   Run  SUCCESS
2  Walk  SUCCESS
3   Run     PASS
4   Run  SUCCESS
5  Walk   FAILED
6  Walk     PASS

In [180]: df_out
Out[180]: 
      FAILED  PASS  SUCCESS
Run        0     1        2
Sit        0     1        0
Walk       1     1        1

在Pandas中使用groupby可以将一列中的内容与另一列进行比较

4 个答案:

时序