我在connect_start
0 2019-01-01 00:01:44
1 2019-01-01 00:02:57
2 2019-01-01 00:24:09
3 2019-01-01 01:35:23
4 2019-01-01 01:46:41
还有advertisement_id
列,客户观看了该列以访问互联网
示例:
0 1
1 2
2 3
3 2
4 1
如何根据value_counts()
或advertisement_id
绘制这两列以查看day
的{{1}}?
我有以下代码:
month
如何对月份进行分组并绘制df = pd.read_csv('./input/data.csv', sep=';')
df['connect_start'] = pd.to_datetime(df['connect_start'], format='%Y/%m/%d %H:%M:%S')
?
这是我的尝试,并且我的计算机多次崩溃。谁能帮忙。
advertisement_id.value()
答案 0 :(得分:0)
您真的很亲密!
将connect_start
列转换为日期时间后,您可以像这样提取month
或day
:
df["my_date_column"].dt.day
然后,您可以像完成操作一样使用groupby
对数据进行分组。
然后,您可以通过调用sum
方法为每个组求和。 (doc)。
最后,您可以从DataFrame.plot.bar
绘制。 (doc)
这是一个有效的示例:
# import modules
import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(2019)
##############################################
# Just for example: create a random dataset #
##############################################
number_row = 10
def random_dates(start, end, n):
""" Generate n random dates between the interval """
start_u = pd.to_datetime(start).value//10**9
end_u = pd.to_datetime(end).value//10**9
return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')
# create the dataframe
df = pd.DataFrame({"connect_start": random_dates("2019-01-01", "2019-12-31", number_row),
"advertisement_id": np.random.randint(0, 10, number_row)})
print(df)
# connect_start advertisement_id
# 0 2019-09-08 18: 34: 48 0
# 1 2019-11-05 06: 30: 10 0
# 2 2019-05-03 01: 32: 15 7
# 3 2019-01-13 06: 37: 25 8
# 4 2019-12-04 03: 47: 36 5
# 5 2019-11-23 14: 12: 14 3
# 6 2019-09-09 18: 39: 50 0
# 7 2019-10-01 08: 38: 53 2
# 8 2019-04-05 21: 37: 19 5
# 9 2019-04-25 04: 26: 52 7
##############################################
# Process #
##############################################
# Convert connect_start as datetime type
df["connect_start"] = pd.to_datetime(df.connect_start, format="%Y-%m-%d %H:%M:%S")
# Extract day of month, date, month in 3 new columns
df["day_in_month"] = df["connect_start"].dt.day
df["day_per_year"] = df['connect_start'].dt.date
df["month"] = df["connect_start"].dt.month
print(df)
# connect_start advertisement_id day_in_month day_per_year month
# 0 2019-09-08 18: 34: 48 0 8 2019-09-08 9
# 1 2019-11-05 06: 30: 10 0 5 2019-11-05 11
# 2 2019-05-03 01: 32: 15 7 3 2019-05-03 5
# 3 2019-01-13 06: 37: 25 8 13 2019-01-13 1
# 4 2019-12-04 03: 47: 36 5 4 2019-12-04 12
# 5 2019-11-23 14: 12: 14 3 23 2019-11-23 11
# 6 2019-09-09 18: 39: 50 0 9 2019-09-09 9
# 7 2019-10-01 08: 38: 53 2 1 2019-10-01 10
# 8 2019-04-05 21: 37: 19 5 5 2019-04-05 4
# 9 2019-04-25 04: 26: 52 7 25 2019-04-25 4
# Compute the number advertisement_id per group
df_day_in_month = df[["advertisement_id", "day_in_month"]].groupby(by="day_in_month").sum()
df_day_per_year = df[["advertisement_id", "day_per_year"]].groupby(by="day_per_year").sum()
df_month = df[["advertisement_id", "month"]].groupby(by="month").sum()
# Fulfill with 0 for all days off the month
df_day_in_month_all_days = df_day_in_month.reindex(np.arange(1, 32), fill_value=0)
#####################################
# Diplay #
#####################################
# Create figure
fig, axes = plt.subplots(nrows=2, ncols=2)
# Define subplots
df_day_in_month.plot.bar(ax=axes[0, 0], rot=0)
df_day_per_year.plot.bar(ax=axes[0, 1], rot=45)
df_day_in_month_all_days.plot.bar(ax=axes[1, 0], rot=0)
df_month.plot.bar(ax=axes[1, 1], rot=0)
# Prevent plot overlapping
plt.tight_layout()
plt.show()
如果您有一些事项的期限不完整(取决于数据集的大小),则可以在没有值的情况下人为地添加0
值。我把它放在左下角。
希望有帮助!