Question

我有两个DataFrame，其中一个包含公交车站号码列表df_stops，另一个包含公交车到港，df_arrivals，（StopNumber和OnTimeStatus = -1，0或1，分别对应于公交车是早期，按时或晚期的。）

我想在df_stops DataFrame中添加3个新列：

PercentEarly
PercentOnTime
PercentLate

我很难在不使用循环迭代的情况下弄清楚如何做到这一点。如果我是迭代地做，我会做一些事情：

for row in df_stops:
    # number of early arrivals / total number of arrivals @ that stop
    row['PercentEarly'] =
        df_arrivals.loc[df_arrivals['StopNum'] == row['StopNum'] and df_arrivals['OnTimeStatus'] < 0].count() 
        / df_arrivals.loc[df_arrivals['StopNum'] == row['StopNum']].count()

    # same idea for on time and late arrivals

我对熊猫和数据科学一般都比较陌生，所以任何帮助都会受到赞赏。

如何在不迭代df_stops的每一行的情况下执行此操作？

编辑：

df_arrivals

       RouteNumber  ScheduledUnix  StopNumber OnTimeStatus
0               44     1511977533       40888            0
1               44     1511979273       40888            0
2               44     1511979273       40888            0
3               44     1511980353       40888            0
4               44     1511979273       40888            0
5               44     1511980353       40888            1
...            ...            ...         ...          ...
67538           85     1512005100       40900            0
67539           85     1512008700       40900            0
67540           85     1512008700       40900           -1
67541           85     1512008700       40900            0
67542           85     1512012300       40900            0

df_stops：

     StopNumber
0         40877
1         40874
2         40876
3         40725
4         40875
5         40776
6         40730
7         40723
8         40721
9         40729
10        40722

所需的输出看起来像：

     StopNumber    EarlyPercent    OnTimePercent    LatePercent
0         40877            0.14             0.80           0.06
...

Answer 1

你可以使用groupby

for stops in df_arrivals.groupby('StopNum'):
    stop[1].groupby('OnTimeStatus').count()

它现在按预期工作吗？

Answer 2

我从未在没有迭代的情况下弄清楚如何做到这一点。我还决定存储早期/准时/晚期的数量而不是百分比。这是我的解决方案，即使给出了成千上万的条目，它似乎也非常快：

chain

Answer 3

回答有关事件发生次数的问题：

我要做的是：

#This represents all early, ontime, and late arrivals. If you want to grab per stopnum then you need to groupby first (see below)
#Define a specific stop num and store as stop_num = the number
early, ontime, late = df_arrivals[df_arrivals.stop_number == stop_num].OnTimeStatus.value_counts()[-1], df_arrivals.OnTimeStatus.value_counts()[0], df_arrivals.OnTimeStatus.value_counts()[1]

total_stops = len(df_stops[df_stops.StopNumber == stop_num])
EarlyPercent= early/total_stops
OntimePercent= ontime/total_stops
LatePercent= late/total_stops

现在请记住，这只是每一站的数量。实际上，我认为在没有过于复杂的代码（链接等等）的情况下，有一种方法可以避免迭代。

df_stops['PercentEarly'] = ''
df_stops['PercentOntime'] = ''
df_stops['PercentLate'] = ''

for stop_num in df_arrivals.stop_number.tolist():
    early, ontime, late = df_arrivals[df_arrivals.stop_number == stop_num].OnTimeStatus.value_counts()[-1], df_arrivals.OnTimeStatus.value_counts()[0], df_arrivals.OnTimeStatus.value_counts()[1]
    total_stops = len(df_stops[df_stops.StopNumber == stop_num])
    EarlyPercent= early/total_stops
    OntimePercent= ontime/total_stops
    LatePercent= late/total_stops
    df_stops.loc[df_stops.StopNumber == stop_num, 'PercentEarly'] =EarlyPercent
    df_stops.loc[df_stops.StopNumber == stop_num, 'PercentOnTime'] = OntimePercent
    df_stops.loc[df_stops.StopNumber == stop_num, 'PercentLate'] =LatePercent

避免迭代以获得熊猫中的出现次数

3 个答案: