我有一系列时间访问建筑物的人。每个人都有一个唯一的身份证。对于时间序列中的每条记录，我想知道在过去365天内访问建筑物的唯一人数（即滚动的唯一计数，窗口为365天）。
pandas似乎没有内置的计算方法。当存在大量唯一访问者和/或大窗口时，计算变得计算密集。 （实际数据大于此示例。）
有没有比我在下面做的更好的计算方法？我不确定为什么我所做的快速方法，windowed_nunique（在“速度测试3”下）是1。
感谢您的帮助！
相关链接：
pandas问题：https://github.com/pandas-dev/pandas/issues/14336 In [1]:
# Import libraries.
import pandas as pd
import numba
import numpy as np
In [2]:
# Create data of people visiting a building.
np.random.seed(seed=0)
dates = pd.date_range(start='2010-01-01', end='2015-01-01', freq='D')
window = 365 # days
num_pids = 100
probs = np.linspace(start=0.001, stop=0.1, num=num_pids)
df = pd\
.DataFrame(
data=[(date, pid)
for (pid, prob) in zip(range(num_pids), probs)
for date in np.compress(np.random.binomial(n=1, p=prob, size=len(dates)), dates)],
columns=['Date', 'PersonId'])\
.sort_values(by='Date')\
.reset_index(drop=True)
print("Created data of people visiting a building:")
df.head() # 9181 rows × 2 columns
Out[2]:
Created data of people visiting a building:
| | Date | PersonId |
|---|------------|----------|
| 0 | 2010-01-01 | 76 |
| 1 | 2010-01-01 | 63 |
| 2 | 2010-01-01 | 89 |
| 3 | 2010-01-01 | 81 |
| 4 | 2010-01-01 | 7 |
In [3]:
%%timeit
# This counts the number of people visiting the building, not the number of unique people.
# Provided as a speed reference.
df.rolling(window='{:d}D'.format(window), on='Date').count()
3.32 ms ± 124 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [4]:
%%timeit
df.rolling(window='{:d}D'.format(window), on='Date').apply(lambda arr: pd.Series(arr).nunique())
2.42 s ± 282 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [5]:
# Save results as a reference to check calculation accuracy.
ref = df.rolling(window='{:d}D'.format(window), on='Date').apply(lambda arr: pd.Series(arr).nunique())['PersonId'].values
In [6]:
# Define a custom function and implement a just-in-time compiler.
@numba.jit(nopython=True)
def nunique(arr):
return len(set(arr))
In [7]:
%%timeit
df.rolling(window='{:d}D'.format(window), on='Date').apply(nunique)
430 ms ± 31.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [8]:
# Check accuracy of results.
test = df.rolling(window='{:d}D'.format(window), on='Date').apply(nunique)['PersonId'].values
assert all(ref == test)
In [9]:
# Define a custom function and implement a just-in-time compiler.
@numba.jit(nopython=True)
def windowed_nunique(dates, pids, window):
r"""Track number of unique persons in window,
reading through arrays only once.
Args:
dates (numpy.ndarray): Array of dates as number of days since epoch.
pids (numpy.ndarray): Array of integer person identifiers.
window (int): Width of window in units of difference of `dates`.
Returns:
ucts (numpy.ndarray): Array of unique counts.
Raises:
AssertionError: Raised if `len(dates) != len(pids)`
Notes:
* May be off by 1 compared to `pandas.core.window.Rolling`
with a time series alias offset.
"""
# Check arguments.
assert dates.shape == pids.shape
# Initialize counters.
idx_min = 0
idx_max = dates.shape[0]
date_min = dates[idx_min]
pid_min = pids[idx_min]
pid_max = np.max(pids)
pid_cts = np.zeros(pid_max, dtype=np.int64)
pid_cts[pid_min] = 1
uct = 1
ucts = np.zeros(idx_max, dtype=np.int64)
ucts[idx_min] = uct
idx = 1
# For each (date, person)...
while idx < idx_max:
# If person count went from 0 to 1, increment unique person count.
date = dates[idx]
pid = pids[idx]
pid_cts[pid] += 1
if pid_cts[pid] == 1:
uct += 1
# For past dates outside of window...
while (date - date_min) > window:
# If person count went from 1 to 0, decrement unique person count.
pid_cts[pid_min] -= 1
if pid_cts[pid_min] == 0:
uct -= 1
idx_min += 1
date_min = dates[idx_min]
pid_min = pids[idx_min]
# Record unique person count.
ucts[idx] = uct
idx += 1
return ucts
In [10]:
# Cast dates to integers.
df['DateEpoch'] = (df['Date'] - pd.to_datetime('1970-01-01'))/pd.to_timedelta(1, unit='D')
df['DateEpoch'] = df['DateEpoch'].astype(int)
In [11]:
%%timeit
windowed_nunique(
dates=df['DateEpoch'].values,
pids=df['PersonId'].values,
window=window)
107 µs ± 63.5 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [12]:
# Check accuracy of results.
test = windowed_nunique(
dates=df['DateEpoch'].values,
pids=df['PersonId'].values,
window=window)
# Note: Method may be off by 1.
assert all(np.isclose(ref, np.asarray(test), atol=1))
In [13]:
# Show where the calculation doesn't match.
print("Where reference ('ref') calculation of number of unique people doesn't match 'test':")
df['ref'] = ref
df['test'] = test
df.loc[df['ref'] != df['test']].head() # 9044 rows × 5 columns
Out[13]:
Where reference ('ref') calculation of number of unique people doesn't match 'test':
| | Date | PersonId | DateEpoch | ref | test |
|----|------------|----------|-----------|------|------|
| 78 | 2010-01-19 | 99 | 14628 | 56.0 | 55 |
| 79 | 2010-01-19 | 96 | 14628 | 56.0 | 55 |
| 80 | 2010-01-19 | 88 | 14628 | 56.0 | 55 |
| 81 | 2010-01-20 | 94 | 14629 | 56.0 | 55 |
| 82 | 2010-01-20 | 48 | 14629 | 57.0 | 56 |
答案 0 :(得分：1)
我在快速方法windowed_nunique中遇到了2个错误，现在已在以下windowed_nunique_corrected中更正：
pid_cts内每个人ID的唯一计数数量的数组的大小太小。 date_min时应更新(date - date_min + 1) > window。相关链接：
In [14]:
# Define a custom function and implement a just-in-time compiler.
@numba.jit(nopython=True)
def windowed_nunique_corrected(dates, pids, window):
r"""Track number of unique persons in window,
reading through arrays only once.
Args:
dates (numpy.ndarray): Array of dates as number of days since epoch.
pids (numpy.ndarray): Array of integer person identifiers.
Required: min(pids) >= 0
window (int): Width of window in units of difference of `dates`.
Required: window >= 1
Returns:
ucts (numpy.ndarray): Array of unique counts.
Raises:
AssertionError: Raised if not...
* len(dates) == len(pids)
* min(pids) >= 0
* window >= 1
Notes:
* Matches `pandas.core.window.Rolling`
with a time series alias offset.
"""
# Check arguments.
assert len(dates) == len(pids)
assert np.min(pids) >= 0
assert window >= 1
# Initialize counters.
idx_min = 0
idx_max = dates.shape[0]
date_min = dates[idx_min]
pid_min = pids[idx_min]
pid_max = np.max(pids) + 1
pid_cts = np.zeros(pid_max, dtype=np.int64)
pid_cts[pid_min] = 1
uct = 1
ucts = np.zeros(idx_max, dtype=np.int64)
ucts[idx_min] = uct
idx = 1
# For each (date, person)...
while idx < idx_max:
# Lookup date, person.
date = dates[idx]
pid = pids[idx]
# If person count went from 0 to 1, increment unique person count.
pid_cts[pid] += 1
if pid_cts[pid] == 1:
uct += 1
# For past dates outside of window...
# Note: If window=3, it includes day0,day1,day2.
while (date - date_min + 1) > window:
# If person count went from 1 to 0, decrement unique person count.
pid_cts[pid_min] -= 1
if pid_cts[pid_min] == 0:
uct -= 1
idx_min += 1
date_min = dates[idx_min]
pid_min = pids[idx_min]
# Record unique person count.
ucts[idx] = uct
idx += 1
return ucts
In [15]:
# Cast dates to integers.
df['DateEpoch'] = (df['Date'] - pd.to_datetime('1970-01-01'))/pd.to_timedelta(1, unit='D')
df['DateEpoch'] = df['DateEpoch'].astype(int)
In [16]:
%%timeit
windowed_nunique_corrected(
dates=df['DateEpoch'].values,
pids=df['PersonId'].values,
window=window)
98.8 µs ± 41.3 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [17]:
# Check accuracy of results.
test = windowed_nunique_corrected(
dates=df['DateEpoch'].values,
pids=df['PersonId'].values,
window=window)
assert all(ref == test)
答案 1 :(得分：0)
如果您只想要在过去365天内进入建筑物内的唯一身份证号码，您可以先使用.loc在过去365天内限制数据集：
df = df.loc[df['date'] > '2016-09-28',:]
和一个群组你得到的行数与进入的独特人物一样多，如果按照计数进行，你也会得到他们进来的次数：
df = df.groupby('PersonID').count()
这似乎适合你的问题，但也许我弄错了。 祝你有个美好的一天
答案 2 :(得分：0)
非常接近您在种子测试2中的时间，但作为一个衬垫，重新采样超过一年。
df.resample('AS',on='Date')['PersonId'].expanding(0).apply(lambda x: np.unique(x).shape[0])
时间结果
1 loop, best of 3: 483 ms per loop