Question

我有一个panda var app = angular.module('MyApp', ['ngRoute']); // configure our routes app.config(function ($routeProvider) { $routeProvider // route for the home page .when('/', { templateUrl: 'AddHeader.html', controller: 'headerCtrl' }) .when('/AddHeader', { templateUrl: 'AddHeader.html', controller: 'headerCtrl' }) // route for the about page .when('/ProjectIDCreation', { templateUrl: 'ProjectIDCreation.html', controller: 'headerCtrl' }) }); var baseAddress = 'http://localhost:49754/api/TimeSheet/'; var url = ""; app.factory('userFactory', function ($http) { return { getHeadersList: function () { url = baseAddress + "FetchHeaderDetails"; return $http.get(url); }, addHeader: function (user) { url = baseAddress + "InsertHeaderDetails"; return $http.post(url, user); }, updateHeader: function (user) { url = baseAddress + "UpdateHeaderDetails"; return $http.put(url, user); } }; }); //var app = angular.module('MyApp'); app.controller('headerCtrl', function PostController($scope, userFactory) { $scope.users = []; $scope.user = null; $scope.editMode = false; //Fetch all Headers $scope.getAll = function () { userFactory.getHeadersList().success(function (data) { $scope.users = data; }).error(function (data) { $scope.error = "An Error has occured while Loading users! " + data.ExceptionMessage; }); }; //Add Header $scope.add = function () { var currentUser = this.user; userFactory.addHeader(currentUser).success(function (data) { $scope.addMode = false; currentUser.HeaderID = data; $scope.users.push(currentUser); $scope.user = null; $('#userModel').modal('hide'); }).error(function (data) { $scope.error = "An Error has occured while Adding user! " + data.ExceptionMessage; }); }; //Edit Header $scope.edit = function () { $scope.user = this.user; $scope.editMode = true; $('#userModel').modal('show'); }; //Update Header $scope.update = function () { var currentUser = this.user; userFactory.updateHeader(currentUser).success(function (data) { currentUser.editMode = false; $('#userModel').modal('hide'); }).error(function (data) { $scope.error = "An Error has occured while Updating user! " + data.ExceptionMessage; }); }; //Model popup events $scope.showadd = function () { $scope.user = null; $scope.editMode = false; $('#userModel').modal('show'); }; $scope.showedit = function () { $('#userModel').modal('show'); }; $scope.cancel = function () { $scope.user = null; $('#userModel').modal('hide'); } // initialize your users data $scope.getAll(); });，时间序列在df，而column1中有一个布尔条件。这描述了满足特定条件的连续时间间隔。请注意，时间间隔长度不等。

column2

如何计算满足此条件的整个系列中的时间间隔总数？

所需的输出应如下所示：

Timestamp   Boolean_condition
1           1
2           1
3           0
4           1
5           1
6           1
7           0
8           0
9           1
10          0

Answer 1

您可以尝试以下方法：

1）获取True实例（此处为1）的所有值，其中包含isone

2）获取相应的索引集并将其转换为系列表示，以便新系列将其索引和值都作为先前计算的索引。执行连续行之间的差异并检查它们是否等于1.这将成为我们的布尔掩码。

3）将isone与获得的布尔掩码进行比较，每当它们不相等时，我们采用它们的累积和（也称为元素之间的邻接检查）。这些帮助我们进行分组。

4）使用loc作为isone的索引，我们将将grp数组更改为分类格式后计算的代码分配给创建的新列 Event_number

isone = df.Bolean_condition[df.Bolean_condition.eq(1)]
idx = isone.index
grp = (isone != idx.to_series().diff().eq(1)).cumsum()
df.loc[idx, 'Event_number'] = pd.Categorical(grp).codes + 1

更快的方法：

仅使用numpy：

1）获取它的数组表示。

2）计算非零，此处（1's）索引。

3）在此数组的开头插入NaN，这将作为我们在考虑连续行时执行差异的起点。

4）初始化一个填充了与原始数组形状相同的Nan's的新数组。

5）每当连续行之间的差异不等于1时，我们取其累积总和，否则它们属于同一组。这些值在之前有1's的指数处被估算。

6）将这些分配回新列。

def nick(df):
    b = df.Bolean_condition.values
    slc = np.flatnonzero(b)
    slc_pl_1 = np.append(np.nan, slc)
    nan_arr = np.full(b.size, fill_value=np.nan)
    nan_arr[slc] = np.cumsum(slc_pl_1[1:] - slc_pl_1[:-1] != 1)
    df['Event_number'] = nan_arr
    return df

<强> 时序：

对于10,000行的DF：

np.random.seed(42)
df1 = pd.DataFrame(dict(
        Timestamp=np.arange(10000),
        Bolean_condition=np.random.choice(np.array([0,1]), 10000, p=[0.4, 0.6]))
                  )

df1.shape
# (10000, 2)

def jez(df):
    mask0 = df.Bolean_condition.eq(0)
    mask2 = df.Bolean_condition.ne(df.Bolean_condition.shift(1))
    df['Event_number'] = (mask2 & mask0).cumsum().mask(mask0)
    return (df)

nick(df1).equals(jez(df1))
# True

%%timeit
nick(df1)
1000 loops, best of 3: 362 µs per loop

%%timeit
jez(df1)
100 loops, best of 3: 1.56 ms per loop

对于包含100万行的DF：

np.random.seed(42)
df1 = pd.DataFrame(dict(
        Timestamp=np.arange(1000000),
        Bolean_condition=np.random.choice(np.array([0,1]), 1000000, p=[0.4, 0.6]))
                  )

df1.shape
# (1000000, 2)

nick(df1).equals(jez(df1))
# True

%%timeit
nick(df1)
10 loops, best of 3: 34.9 ms per loop

%%timeit
jez(df1)
10 loops, best of 3: 50.1 ms per loop

Answer 2

您可以使用cumsum两个Series创建masks，然后按功能Series.mask创建NaN：

mask0 = df.Boolean_condition.eq(0)
mask2 = df.Boolean_condition.ne(df.Boolean_condition.shift(1))
print ((mask2 & mask0).cumsum().add(1))
0    1
1    1
2    2
3    2
4    2
5    2
6    3
7    3
8    3
9    4
Name: Boolean_condition, dtype: int32

df['Event_number'] = (mask2 & mask0).cumsum().add(1).mask(mask0)
print (df)
   Timestamp  Boolean_condition  Event_number
0          1                  1           1.0
1          2                  1           1.0
2          3                  0           NaN
3          4                  1           2.0
4          5                  1           2.0
5          6                  1           2.0
6          7                  0           NaN
7          8                  0           NaN
8          9                  1           3.0
9         10                  0           NaN

<强>计时：

#[100000 rows x 2 columns
df = pd.concat([df]*10000).reset_index(drop=True)
df1 = df.copy()
df2 = df.copy()

def nick(df):
    isone = df.Boolean_condition[df.Boolean_condition.eq(1)]
    idx = isone.index
    grp = (isone != idx.to_series().diff().eq(1)).cumsum()
    df.loc[idx, 'Event_number'] = pd.Categorical(grp).codes + 1
    return df

def jez(df):
    mask0 = df.Boolean_condition.eq(0)
    mask2 = df.Boolean_condition.ne(df.Boolean_condition.shift(1))
    df['Event_number'] = (mask2 & mask0).cumsum().add(1).mask(mask0)
    return (df)

def jez1(df):
    mask0 = ~df.Boolean_condition
    mask2 = df.Boolean_condition.ne(df.Boolean_condition.shift(1))
    df['Event_number'] = (mask2 & mask0).cumsum().add(1).mask(mask0)
    return (df)

In [68]: %timeit (jez1(df))
100 loops, best of 3: 6.45 ms per loop

In [69]: %timeit (nick(df1))
100 loops, best of 3: 12 ms per loop

In [70]: %timeit (jez(df2))
100 loops, best of 3: 5.34 ms per loop

Answer 3

自定义功能可以解决问题。这是Matlab代码中的解决方案：

Boolean_condition = [1 1 0 1 1 1 0 0 1 0];
Event_number = [NA NA NA NA NA NA NA NA NA NA];
loop_event_number = 1;

for timestamp=1:10
if Boolean_condition(timestamp)==1
Event_number(timestamp) = loop_event_number;
last_event_number = loop_event_number;
else
loop_event_number = last_event_number +1;
end
end

% Event_number = 1 1 NA 2 2 2 NA NA 3 NA

Answer 4

这应该可以工作，但对于很长的df可能有点慢。

df = pd.concat([df,pd.Series([0]*len(df), name = '2')], axis = 1)
if df.iloc[0,1] == 1:
        counter = 1
        df.iloc[0, 2] = counter
    else:
        counter = 0
        df.iloc[0,2] = 0
    previous = df.iloc[0,1]
    for y,x in df.iloc[1:,].iterrows():
        print(y)
        if x[1] == 1 and previous == 1:
            previous = x[1]
            df.iloc[y, 2] = counter
        if x[1] == 0:
            previous = x[1]
            df.iloc[y,2] = 0
        if x[1] == 1 and previous == 0:
            counter += 1
            previous = x[1]
            df.iloc[y,2] = counter

如何计算在pandas数据帧中满足布尔条件的时间间隔数？

4 个答案: