基于用户名确定两个状态之间的时间差

Question

我试图计算应用程序上用户名状态从原始安装/重新安装到卸载（反之亦然）之间的时间。我已经用Python编程大约一年了，所以我的函数有时较慢或不是最佳的；我在这项任务中遇到了类似的问题

我使用下面的循环来计算状态之间的时间差。主要的问题是这个函数速度非常慢（对于我正在处理的700k行数据），并且它不会返回所有的时间间隔

# Import modules
import pandas as pd
import numpy as np

# Load sample data
data = pd.read_csv('username_status_dates.csv').drop(columns='index')
data['Date'] = pd.to_datetime(data['Date'], format='%Y-%M-%d')

# Create dataframe to record time intervals
customerStatuses_df = pd.DataFrame(columns=['username', 'status1', 'status2', 'timeToUninstall', 'timeToReinstall'])

# Iterate over data, calculate time intervals (by username), and add to customerStatuses
for user in data:

    # Get user sub data
    user_df = data.loc[ customerStatuses_df['Username'] == user ]
    dates = list(user_df['Date'])
    status = list(user_df['Status'])

    for val in np.arange(0, len(status) - 1):

        # If status in installed (original or re)
        if (status[val] == 'Original Install') or (status[val] == 'Reinstall'):

            # If next status is uninstall (un)
            if status[val + 1] == 'Uninstall':

                # Calculate time interval between statuses
                d_unin = dates[val + 1] - dates[val]

                # Add to customerStatuses
                customerStatuses_df = customerStatuses_df.append({'username': user, 'status1': status[val], 'status2': status[val+1], 'timeToUninstall': d_unin}, ignore_index=True)

        # If status in not installed (un)      
        elif (status[val] == 'Uninstall'):

            # If next status is installed (re)
            if status[val + 1] == 'Reinstall':

                # Calculate time interval between statuses
                d_rein = dates[val + 1] - dates[val]

                # Add to customerStatuses
                 customerStatuses_df = customerStatuses_df.append({'username': user, 'status1': status[val], 'status2': status[val+1], 'timeToReinstall': d_rein}, ignore_index=True)

示例数据：username\u status\u dates.csv

   Username         Status                  Date
0   Joey74      Original Install    2019-01-20 00:09:00
1   Sam22       Original Install    2019-01-15 00:10:00
2   Chloe35     Uninstall           2019-01-30 00:10:00
3   Sam22       Uninstall           2019-01-01 00:11:00
4   Zoe44       Reinstall           2019-01-10 00:12:00
5   Joey74      Reinstall           2019-01-20 00:12:00
6   Shelby99    Original Install    2020-01-07 00:01:00
7   Joey74      Uninstall           2020-01-05 00:01:00
8   Joey74      Reinstall           2020-01-30 00:01:00
9   Susane11    Reinstall           2020-01-04 00:02:00

函数的结果：客户状态

username    status1 status2 timeToUninstall timeToReinstall
0   Joey74  Reinstall   Uninstall   349 days 23:49:00   NaN
1   Joey74  Uninstall   Reinstall   NaT 25 days 00:00:00
2   Sam22   Original Install    Uninstall   -14 days +00:01:00  NaN

我错过了什么，在这里

Answer 1

我和一位同事意识到，这个循环的核心问题是它试图迭代数据帧本身，与其他梳理数据帧的方法相比，它的速度异常缓慢。这部分解释了为什么我的函数（上面）在使用更大的数据集时速度如此之慢。我们还发现使用矢量化数字比使用日期列的timedelta格式快得多

我们发现了一个很棒的StackOverflow source，它展示了如何将数据帧转换（或内爆）为字典，以取代对数据帧本身进行迭代的需要为了回答上述问题，我按用户名组织数据，对日期列进行矢量化，并将数据框转换为字典，然后计算状态更改之间的时间：

# Sort by username, then date
customerStatuses_df = customerStatuses_df.sort_values(by=['Username', 'Date'])

# Vectorize the Date column with .diff()
vectorized_time = customerStatuses_df['Date'].diff()
customerStatuses_df['vectorizedTime'] = pd.Series(vectorized_time)

# Implode the dataframe by making it a dictionary
customerStatuses_dict = customerStatuses_df.to_dict('index')

# Create lists to form a final dataframe
username = []
status_1 = []
status_2 = []
time_change = []

# Track timing of function
from tqdm import tqdm

# Determine difference between states, by user
imploded_dict = {}

for key, value in tqdm(list(customerStatuses_dict.items())):

    # Grab all install Statuses and vectorizedTime attributed to the Username 'key'
    try:
        # Python prefers 'ask forgiveness, not permission' -> attempt to append to list
        imploded_dict[value['Username']].append((value['Status'], value['vectorizedTime']))

    except KeyError:  # but if list does not exist, create the list
        imploded_dict[value['Username']] = [(value['Status'], value['vectorizedTime'])]

# Using the now-imploded data, calculate time change between states
for user_id, status_tuple_list in tqdm(list(imploded_dict.items())):

    # Iterate key value pairs in dictionary
    for idx, (status, time_delta) in enumerate(status_tuple_list): 
        if idx == 0:
            continue

        # Add to lists for final dataframe
        username.append(user_id)
        status_1.append(status_tuple_list[idx - 1][0])
        status_2.append(status_tuple_list[idx][0])
        time_change.append(time_delta)


# Create dataframe to hold time changes
results_df = pd.DataFrame(list(zip(username, status_1, status_2, time_change)), columns=['username', 'status1', 'status2', 'timeToChange'])

结果\u df

    username     status1             status2            timeToChange
0   Joey74      Original Install    Reinstall          0 days 00:03:00
1   Joey74      Reinstall           Uninstall          349 days 23:49:00
2   Joey74      Uninstall           Reinstall          25 days 00:00:00
3   Sam22       Uninstall           Original Install   13 days 23:59:00