英文:
How to divide a dataframe by the number of days in a month?
问题
我理解，你想要一个新的数据框，将原始数据框中的数值除以每个月的天数。你的代码中似乎有一些HTML编码和未定义的变量，但我可以提供一个简化版本的代码来执行这个任务：
import pandas as pd# 假设你已经有了原始数据框 df# 将 "Time" 列转换为日期时间类型df['Time'] = pd.to_datetime(df['Time'])# 计算每个月的天数df['Days_in_Month'] = df['Time'].dt.daysinmonth# 将 var1 和 var2 列除以每个月的天数df['var1_new'] = df['var1'] / df['Days_in_Month']df['var2_new'] = df['var2'] / df['Days_in_Month']# 删除中间列 "Days_in_Month"df = df.drop('Days_in_Month', axis=1)
这个代码会为你创建一个新的数据框，其中包含了 "var1_new" 和 "var2_new" 列，它们是原始数据除以每个月天数后的结果。
英文:
I have a Pandas data frame that looks like this below:
var1 var2Time2000-01-31 100 2002000-02-28 340 2102000-03-31 590 220...2001-10-31 1290 1012001-11-30 1188 1002001-12-31 1000 100
I would like to create a new data frame that divides the first data frame by the number of days in each month. For instances,
2000-01-31, divide by 31.
2000-02-28, divide by 28.
2001-11-30, divide by 30.
2001-12-31, divide by 31.
And so on. The new data frame would look something like this
var1_new var2_newTime2000-01-31 3.22 6.452000-02-28 12.14 7.52000-03-31 19.03 7.09...2001-10-31 41.61 3.262001-11-30 39.6 3.332001-12-31 32.26 3.23
I constructed a small little code where I first used a timedelta to index the data frame. I took the sum of each day and month. I then created another data frame that records the number of days in a month. Finally, I tried to divide the first data frame but the number of days in a month to get the above results. However, all I get is a bunch of NaN which I have no idea why.
df.index = pd.to_timedelta(df["Time"], unit='s')df.index += datetime.strptime(initial_datetime, "%Y-%m-%d")df_D=df.resample("D").sum()df_M=df_D.resample("M").sum()df_M["Months"] = df_M.indexdf2["DiM"] = pd.DataFrame(pd.to_datetime(df_M["Months"]).dt.daysinmonth)df_M = df_M.drop(['Months'],axis=1)df_New = df_M / df2["DiM"]
答案1
得分: 1
假设您的索引（Time）已经是类型为pd.Timestamp：
df.apply(lambda col: col / df.index.days_in_month)
英文:
Assuming your index (Time) is already of type pd.Timestamp:
df.apply(lambda col: col / df.index.days_in_month)
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