Question

我正在学习Hadly的书“ R for Data Science”，并试图在熊猫中复制代码。

我遇到了这个问题：
我必须根据延迟时间创建一个新的等级列排期并仅过滤它们的最小值和最大值。

R码：

library(nycflights13)
library(dplyr)

# remove nans
not_cancelled = flights %>% 
    filter( !is.na(dep_delay), !is.na(arr_delay))

# create new column of rank based on dep_time for each day.
df = not_cancelled %>%  
    group_by(year,month,day) %>%  
    mutate(r = min_rank(desc(dep_time))) %>%  
    filter(r %in% range(r)) %>%  # filter only first and last value
    select(year,month,day,dep_delay,arr_delay,r)

dim(df)
head(df,10)

这给出了：

m=month d =day dl = dep_delay ad = arr_delay r =r
    year    m   d   dl  ad  r
    2013    1   1    2  11  831
    2013    1   1   -3  -12 1
    2013    1   2   43  36  928
    2013    1   2   -5  -24 1
    2013    1   3   33  22  900
    2013    1   3   -10 -11 1
    2013    1   4   26  23  908
    2013    1   4   -1  -8  1 
    2013    1   4   -1  -9  1 # Behold! january 4 has 3 rows!!
    2013    1   5   15  18  717

我正试图在大熊猫中复制它：

df = pd.read_csv('https://github.com/bhishanpdl/Datasets/blob/master/nycflights13.csv?raw=true')
# print(df.shape)
# print(df.iloc[:5,:5])

not_cancelled = df.dropna(subset=['dep_delay','arr_delay'])

df['r'] = not_cancelled.groupby(['year','month','day'])['dep_time']\
    .rank('min',ascending=False)

g = df.groupby(['year','month','day'])['r']
g = g.agg([min,max]).reset_index()

f = g.head()
print(f)

Python输出：

(336776, 19)
   year  month  day  min    max
0  2013      1    1  1.0  831.0
1  2013      1    2  1.0  928.0
2  2013      1    3  1.0  900.0
3  2013      1    4  1.0  908.0
4  2013      1    5  1.0  717.0

这不太正确。怎么做正确的事？

我们非常感谢您的帮助。冰雹大熊猫！

Answer 1

这是正确的输出，您只需要调整输出的形状

方法1 stack

g = df.groupby(['year','month','day'])['r']
g = g.agg([min,max]).stack()
g=g.reset_index(level=[0,1,2])

方法2 melt

g=df.groupby(['year','month','day'])['r'].agg([min,max])
g.reset_index().melt(['year','month','day'])

更新

g = df.groupby(['year','month','day'])['r']
g_max = g.transform('max')
g_min = g.transform('min')
yourdf=df.loc[(df.r==g_max)|(df.r==g_min),['year','month','day','r']]

Answer 2

我创建了两个等级，其中最大值为1，最小值为1。

然后我可以获得最大或最小排名为1的行。

但这给了我两列-一列为r_max，一列为r_min

import pandas as pd

df = pd.read_csv('https://github.com/bhishanpdl/Datasets/blob/master/nycflights13.csv?raw=true')
# print(df.shape)
# print(df.iloc[:5,:5])

not_cancelled = df.dropna(subset=['dep_delay','arr_delay'])

gr = not_cancelled.groupby(['year','month','day'])

df['r_min'] = gr['dep_time'].rank('min', ascending=False)
df['r_max'] = gr['dep_time'].rank('max', ascending=True)

result = df[(df['r_min'] == 1) | (df['r_max'] == 1)]

print(result[['year','month','day','dep_delay','arr_delay','r_min', 'r_max']].head(10))

结果-january 4的三行

      year  month  day  dep_delay  arr_delay  r_min  r_max
0     2013      1    1        2.0       11.0  831.0    1.0
837   2013      1    1       -3.0      -12.0    1.0  831.0
842   2013      1    2       43.0       36.0  928.0    1.0
1776  2013      1    2       -5.0      -24.0    1.0  928.0
1785  2013      1    3       33.0       22.0  900.0    1.0
2688  2013      1    3      -10.0      -11.0    1.0  900.0
2699  2013      1    4       26.0       23.0  908.0    1.0
3606  2013      1    4       -1.0       -8.0    1.0  908.0
3607  2013      1    4       -1.0       -9.0    1.0  908.0
3614  2013      1    5       15.0       18.0  717.0    1.0

在pandas groupby之后仅过滤组的几个元素

2 个答案: