Question

具有这样的数据框-

df = {'Request': [0, 0, 1, 0, 1, 0, 0],
 'Time': ['16:00', '17:00', '18:00', '19:00', '20:00', '20:30', '24:00'],
 'grant': [3, 0, 0, 5, 0, 0, 5]}

pd.DataFrame(df).set_index('Time')

    Out[16]: 
       Request  grant
Time                 
16:00        0      3
17:00        0      0
18:00        1      0
19:00        0      5
20:00        1      0
20:30        0      0
24:00        0      5

“请求”列中的值是布尔值，表示是否发出了请求。 1 =请求0 =无请求。 “赠款”列中的值表示初始赠款额。

我想为每个请求计算请求和授予之间的时间。因此，在这种情况下，他们将是19:00-18:00 = 1小时和24：00-20：00 = 4小时。有没有办法使用熊猫轻松地对大型数据集执行此操作？

Answer 1

我会这样处理：

df = {'Request': [0, 0, 1, 0, 1, 0, 0],
     'Time': ['16:00', '17:00', '18:00', '19:00', '20:00', '20:30', '24:00'],
     'grant': [3, 0, 0, 5, 0, 0, 5]}

df = pd.DataFrame(df) #create DataFrame

#get rid of any rows have neither a grant nor request
df = df[(df[['grant', 'Request']].T != 0).any()] 

#change the time in HH:MM to number of minutes
df['Time'] = df['Time'].str.split(":").apply(lambda x: int(x[0])*60 + int(x[1]))

#get the difference between those times
df['timeElapsed'] = df['Time'].diff()

#filter out the requests to only get the grants and their times. 
#Also, drop the NA from the first line.
df = df[(df[['grant']].T != 0).any()].dropna()

#drop all columns except timeElapsed and Grant
df = df[['timeElapsed', 'grant']]

然后输出类似于timeElaped（以分钟为单位）：

   timeElapsed  grant
3         60.0      5
6        240.0      5

Answer 2

首先，您需要将Can't parse '2018.000000106' as date with format 'YYYYMMDD'索引转换为可减去的值才能找到时间增量。因为没有Time，所以无法使用pd.to_timestamp。下面的解决方案使用十进制时间（1:30 PM = 13.5）：

24:00

结果：

# Convert the index into decimal time
df.index = pd.to_timedelta(df.index + ':00') / pd.Timedelta(hours=1)

# Get time when each request was made
r = df[df['Request'] != 0].index.to_series()

# Get time where each grant was made
g = df[df['grant'] != 0].index.to_series()

# `asof` mean "get the last available value in `r` as the in `g.index`
tmp = r.asof(g)
df['Delta'] = tmp.index - tmp

根据不同列的值对数据框执行查找

2 个答案: