我有以下 Pandas 數(shù)據(jù)框，其中時(shí)間（持續(xù)時(shí)間）以一種非常奇怪的格式給出：Person Activity Duration1 A 1 00:002 A 1 00:003 B 0 21:174 C 0 17:11其中1 00:00表示 24 小時(shí)，0 21:17表示 0 天和 21:17 小時(shí)，也就是說只有 21:17 小時(shí)。快速查看 dtypes 返回：In[1]: df.dtypesOut[1]: Person objectActivity objectDuration objectdtype: object如果值為 ，我如何始終如一地對(duì)待該Duration列以返回 24 1 00:00，如果我有，則返回持續(xù)時(shí)間的十進(jìn)制值0 21:17？的十進(jìn)制值為0 21:1721.283。結(jié)果應(yīng)該是：Person Activity Duration1 A 242 A 243 B 21.2834 C 17.183
5 回答
慕少森
TA貢獻(xiàn)2019條經(jīng)驗(yàn) 獲得超9個(gè)贊
a = np.array([24, 1, 1/60])
df.Duration = df.Duration.str.split(' |:', expand=True).astype(int).dot(a)
例子：
df = pd.DataFrame({'Person': [1,2,3,4], "Activity": list('AABC') ,"Duration":['1 00:00', '1 00:00', '0 21:17', '0 17:11']})
df.Duration = df.Duration.str.split(' |:', expand=True).astype(int).dot(a)
print(df)
# Person Activity Duration
#0 1 A 24.000000
#1 2 A 24.000000
#2 3 B 21.283333
#3 4 C 17.183333
森欄
TA貢獻(xiàn)1810條經(jīng)驗(yàn) 獲得超5個(gè)贊
正如您提到的，它不會(huì)超過 1 00:00，即 24:00，有一種更簡(jiǎn)單的方法：
'''
Person Activity Duration
1 A 1 00:00
2 A 1 00:00
3 B 0 21:17
4 C 0 17:11
'''
import pandas as pd
df = pd.read_clipboard("\t")
.
Person Activity Duration
0 1 A 1 00:00
1 2 A 1 00:00
2 3 B 0 21:17
3 4 C 0 17:11
.
df['Duration'] = df['Duration'].str.split(' ')
df['Duration'] = ['24:00' if int(val[0]) == 1 else val[1] for val in df['Duration']]
print(df)
.
Person Activity Duration
0 1 A 24:00
1 2 A 24:00
2 3 B 21:17
3 4 C 17:11
繁花不似錦
TA貢獻(xiàn)1851條經(jīng)驗(yàn) 獲得超4個(gè)贊
您可以輕松地將這些數(shù)字相乘和相加：
durations = [
"1 00:00",
"0 21:17",
]
for duration in durations:
day, clock = duration.split()
hour, minute = clock.split(':')
print((int(day) * 24) + int(hour) + (int(minute) / 60))
青春有我
TA貢獻(xiàn)1784條經(jīng)驗(yàn) 獲得超8個(gè)贊
您可以使用 datetime 模塊進(jìn)行時(shí)間轉(zhuǎn)換
from datetime import datetime
def durationInDecimal(string):
day, time = string.split(" ")
t = datetime.strptime(time, "%H:%M").time()
return int(day)*24 + (t.hour+t.minute/60.0)
df = pd.DataFrame({'Person': list("ABCD"), "Activity": list('ABCD') ,"duration":['1 00:00', '1 00:00', '0 21:17', '0 17:11']})
df["duration"] = df.duration.apply(durationInDecimal)
# Person Activity duration
# 0 A A 24.000000
# 1 B B 24.000000
# 2 C C 21.283333
# 3 D D 17.183333
明月笑刀無情
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超4個(gè)贊
除了其他有用的答案之外，我還想發(fā)布我自己的解決方案，它使用自定義函數(shù)并將其應(yīng)用于數(shù)據(jù)框df.apply：
def custom_time_to_decimals(value):
if value.split()[0]=='1':
return 24
else:
custom = value.split()[1]
hours = int(custom[0:2])
minutes = int(custom[3:5])
decimal = hours + (minutes/60)
return round(decimal,3)
df['decimalHours'] = df['<insertYourTimeColumnHere>'].apply(custom_time_to_decimals)
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