我有以下熊貓df（datetime屬于類型datetime64）： device datetime0 846ee 2020-03-22 14:27:291 0a26e 2020-03-22 15:33:312 8a906 2020-03-27 16:19:063 6bf11 2020-03-27 16:05:204 d3923 2020-03-23 18:58:51我想使用 Seaborn 的 KDE 功能distplot。盡管我不完全明白為什么，但我還是讓它工作了：df['hour'] = df['datetime'].dt.floor('T').dt.timedf['hour'] = pd.to_timedelta(df['hour'].astype(str)) / pd.Timedelta(hours=1)進而sns.distplot(df['hour'], hist=False, bins=arr, label='tef')問題是：我如何做同樣的事情，但只計算 unique devices？我努力了df.groupby(['hour']).nunique().reset_index()df.groupby(['hour'])[['device']].size().reset_index()但是他們給了我不同的結(jié)果（數(shù)量級相同，但或多或少）。我想我不明白我在做什么pd.to_timedelta(df['hour'].astype(str)) / pd.Timedelta(hours=1)，這讓我無法思考獨特之處……也許吧。
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TA貢獻1828條經(jīng)驗 獲得超6個贊
pd.to_timedelta(df['time'].astype(str))箱子輸出像0 days 01:00:00pd.to_timedelta(df['time'].astype(str)) / pd.Timedelta(hours=1)創(chuàng)建類似 的輸出1.00，它是float小時數(shù)。timedeltas。
import pandas as pd
import numpy as np? # for test data
import random? # for test data
# test data
np.random.seed(365)
random.seed(365)
rows = 40
data = {'device': [random.choice(['846ee', '0a26e', '8a906', '6bf11', 'd3923']) for _ in range(rows)],
? ? ? ? 'datetime': pd.bdate_range(datetime(2020, 7, 1), freq='15min', periods=rows).tolist()}
# create test dataframe
df = pd.DataFrame(data)
# this date column is already in a datetime format; for the real dataframe, make sure it's converted
# df.datetime = pd.to_datetime(df.datetime)
# this extracts the time component from the datetime and is a datetime.time object
df['time'] = df['datetime'].dt.floor('T').dt.time
# this creates a timedelta column; note it's format
df['timedelta'] = pd.to_timedelta(df['time'].astype(str))
# this creates a float representing the hour and its fractional component (minutes)
df['hours'] = pd.to_timedelta(df['time'].astype(str)) / pd.Timedelta(hours=1)
# extracts just the hour
df['hour'] = df['datetime'].dt.hour
顯示（df.head（））
這個觀點應(yīng)該闡明時間提取方法之間的區(qū)別。
? ?device? ? ? ? ? ? datetime? ? ? time? ? ? ?timedelta? hours? hour
0? ?8a906 2020-07-01 00:00:00? 00:00:00 0 days 00:00:00? ?0.00? ? ?0
1? ?0a26e 2020-07-01 00:15:00? 00:15:00 0 days 00:15:00? ?0.25? ? ?0
2? ?8a906 2020-07-01 00:30:00? 00:30:00 0 days 00:30:00? ?0.50? ? ?0
3? ?d3923 2020-07-01 00:45:00? 00:45:00 0 days 00:45:00? ?0.75? ? ?0
4? ?0a26e 2020-07-01 01:00:00? 01:00:00 0 days 01:00:00? ?1.00? ? ?1
5? ?d3923 2020-07-01 01:15:00? 01:15:00 0 days 01:15:00? ?1.25? ? ?1
6? ?6bf11 2020-07-01 01:30:00? 01:30:00 0 days 01:30:00? ?1.50? ? ?1
7? ?d3923 2020-07-01 01:45:00? 01:45:00 0 days 01:45:00? ?1.75? ? ?1
8? ?6bf11 2020-07-01 02:00:00? 02:00:00 0 days 02:00:00? ?2.00? ? ?2
9? ?d3923 2020-07-01 02:15:00? 02:15:00 0 days 02:15:00? ?2.25? ? ?2
10? 0a26e 2020-07-01 02:30:00? 02:30:00 0 days 02:30:00? ?2.50? ? ?2
11? 846ee 2020-07-01 02:45:00? 02:45:00 0 days 02:45:00? ?2.75? ? ?2
12? 0a26e 2020-07-01 03:00:00? 03:00:00 0 days 03:00:00? ?3.00? ? ?3
13? 846ee 2020-07-01 03:15:00? 03:15:00 0 days 03:15:00? ?3.25? ? ?3
14? 846ee 2020-07-01 03:30:00? 03:30:00 0 days 03:30:00? ?3.50? ? ?3
繪制每小時的設(shè)備計數(shù)seaborn.countplot
plt.figure(figsize=(8, 6))
sns.countplot(x='hour', hue='device', data=df)
plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left')
seaborn.distplot為每個設(shè)備繪制 a
使用
seaborn.FacetGrid這將給出每個設(shè)備的每小時分布。
import seaborn as sns
import matplotlib.pyplot as plt
g = sns.FacetGrid(df, row='device', height=5)
g.map(sns.distplot, 'hours', bins=24, kde=True)
g.set(xlim=(0, 24), xticks=range(0, 25, 1))
婷婷同學(xué)_
TA貢獻1844條經(jīng)驗 獲得超8個贊
你可以試試
df['hour'] = df['datetime'].dt.strftime('%Y-%m-%d %H')
s = df.groupby('hour')['device'].value_counts()添加回答
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