如果单独的每小时日期时间列位于指定的小时范围内，则创建新的对象列值。

Question

我有一列'Flight_Time‘，它显示飞行的时间和分钟(%H:%M格式，例如02:22，timedelta64ns)，我想创建一个新列(df’‘val’)，将飞行时间分类为‘0-2小时’、‘2-4小时’、‘4-6小时’或'6+hrs‘，这样我就可以画出这4个新变量的总和。

有人能建议我如何设置下面的if语句来创建“val”列来对这4个子集进行分类吗？飞行时间列是一个timedelta64ns对象。

if df['Flight_Time'] >= '0:00' & df['Flight_Time'] < '02:00':
    df['val'] = '0-2 hrs'
elif df['Flight_Time'] >= '2:00' & df['Flight_Time'] < '04:00':
    df['val'] = '2-4 hrs'
elif df['Flight_Time'] >= '4:00' & df['Flight_Time'] < '06:00':
    df['val'] = '4-6 hrs'
else:
    df['val'] = '6+ hrs' 

目标dataframe输出将类似于：

Flight Time  val
0   00:00   0-2 hr
1   00:01   0-2 hr
2   04:05   2-4 hr
3   10:08   6+ hr
4   02:10   2-4 hr

更新:我已经将代码更改为下面的代码，但是现在我只在新创建的'Val‘列中获得6+小时

from datetime import timedelta

for x in df["Flight_Time"]:
    if  timedelta(hours = 0, minutes = 0) < x <= timedelta(hours = 2, minutes = 0):
        df['val'] = '0-2 hrs'
    elif timedelta(hours = 2, minutes = 0) < x <= timedelta(hours = 4, minutes = 0):
        df['val'] = '2-4 hrs'
    elif timedelta(hours = 4, minutes = 0) < x <= timedelta(hours = 6, minutes = 0):
        df['val'] = '4-6 hrs'
    else:
        df['val'] = '6+ hrs' 

Answer 1

"""it's a bit redundant as implementation
however it shows how to handle
rows and columns with pandas the way you want
- just add the comparison using datetime types
(because here I used integers)
"""

import pandas as pd

df = pd.DataFrame([
    [0.5],
    [3],
    [5],
    [10]
], columns=["flight_time"])


df["flight_interval"] = None
df.loc[df["flight_time"] < 2, ["flight_interval"]] = "0-2 hrs"
df.loc[(df["flight_time"] > 2) & (df["flight_time"] < 4), ["flight_interval"]] = "2-4 hrs"
df.loc[(df["flight_time"] > 4) & (df["flight_time"] < 6), ["flight_interval"]] = "4-6 hrs"
df.loc[(df["flight_time"] > 6) , ["flight_interval"]] = "6+ hrs"
print(df)

输出

+----+---------------+-------------------+
|    |   flight_time | flight_interval   |
|----+---------------+-------------------|
|  0 |           0.5 | 0-2 hrs           |
|  1 |           3   | 2-4 hrs           |
|  2 |           5   | 4-6 hrs           |
|  3 |          10   | 6+ hrs            |
+----+---------------+-------------------+

-编辑:具有timedelta类型的版本

import pandas as pd
from datetime import timedelta

df = pd.DataFrame([
    [0.5],
    [3],
    [5],
    [10]
], columns=["flight_time"])
df["flight_time"] = pd.to_timedelta(df["flight_time"], unit="hours")

df["flight_interval"] = None
df.loc[df["flight_time"] < timedelta(hours=2), ["flight_interval"]] = "0-2 hrs"
df.loc[(df["flight_time"] > timedelta(hours=2)) & (df["flight_time"] < timedelta(hours=4)), ["flight_interval"]] = "2-4 hrs"
df.loc[(df["flight_time"] > timedelta(hours=4)) & (df["flight_time"] < timedelta(hours=6)), ["flight_interval"]] = "4-6 hrs"
df.loc[(df["flight_time"] > timedelta(hours=6)), ["flight_interval"]] = "6+ hrs"

+----+-----------------+-------------------+
|    | flight_time     | flight_interval   |
|----+-----------------+-------------------|
|  0 | 0 days 00:30:00 | 0-2 hrs           |
|  1 | 0 days 03:00:00 | 2-4 hrs           |
|  2 | 0 days 05:00:00 | 4-6 hrs           |
|  3 | 0 days 10:00:00 | 6+ hrs            |
+----+-----------------+-------------------+

Answer 2

像这样的事情应该有效：

df["val"] = pd.cut(df["Flight_Time"], bins=[2,4,6,8,10,12])