获取一组字符串中的最大值和最小值

values = [5, 6,7,8 , 9, 11,12, 13, 14, 17, 18,19, 20, 21,22, 23, 
          24, 25, 26, 27, 41, 42, 44, 45, 46, 47]
s = pd.Series(values)
s1 = s.groupby(s.diff().gt(1).cumsum()).apply(lambda x: ','.join(x.astype(str)))
print (s1)

0: 5,6,7,8,9

1: 11,12,13,14

2: 17,18,19,20,21,22,23,24,25,26,27

3: 41,42

4: 44,45,46,47

I am trying to find the min and max of each line of the group. I have tried several approaches, but I don't get it correctly.

我试图找到该组每行的最小值和最大值。我尝试了几种方法,但我没有正确理解。

My belief is, it has to be converted to int, then the maximum and minimum can be found, but I am not sure how to do that. Every time I try to access the series it converts to strings.

我的信念是,它必须转换为int,然后可以找到最大值和最小值,但我不知道该怎么做。每次我尝试访问该系列时,它都会转换为字符串。

The output will be in a form of min and max value in the following for loop:

以下for循环中的输出将采用最小值和最大值的形式:

for num in s1:
    min_value = 
    max_value = 
    print(min_value ,max_value )

4 个解决方案

#1

I suggest create lists instead joined strings and then use min and max:

我建议创建列表而不是连接字符串,然后使用min和max:

s1 = s.groupby(s.diff().gt(1).cumsum()).apply(list)
print (s1)
0                                 [5, 6, 7, 8, 9]
1                                [11, 12, 13, 14]
2    [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]
3                                        [41, 42]
4                                [44, 45, 46, 47]
dtype: object

for num in s1:
    min_value = min(num)
    max_value = max(num)
    print(min_value ,max_value)

Or better use groupby object and first join to strings and then aggregate min and max:

或者更好地使用groupby对象并首先连接到字符串然后聚合min和max:

g = s.groupby(s.diff().gt(1).cumsum())
s1 = g.apply(lambda x: ','.join(x.astype(str)))
print (s1)
0                           5,6,7,8,9
1                         11,12,13,14
2    17,18,19,20,21,22,23,24,25,26,27
3                               41,42
4                         44,45,46,47
dtype: object

s1 = g.agg([min, max])
print (s1)
   min  max
0    5    9
1   11   14
2   17   27
3   41   42
4   44   47

But if need working with joined strings is possible split and convert to int, last get min and max:

但是如果需要使用连接字符串可以拆分并转换为int,则最后得到min和max:

s1 = s.groupby(s.diff().gt(1).cumsum()).apply(lambda x: ','.join(x.astype(str)))
print (s1)
0                           5,6,7,8,9
1                         11,12,13,14
2    17,18,19,20,21,22,23,24,25,26,27
3                               41,42
4                         44,45,46,47
dtype: object

for line in s1:
    a = [int(x) for x in line.split(',')]
    min_value = min(a)
    max_value = max(a)
    print(min_value ,max_value)

#2

One suggestion:

import pandas as pd

values = [5, 6,7,8 , 9, 11,12, 13, 14, 17, 18,19, 20, 21,22, 23, 
          24, 25, 26, 27, 41, 42, 44, 45, 46, 47]
s = pd.Series(values)
s1 = s.groupby(s.diff().gt(1).cumsum()).apply(lambda x: ','.join(x.astype(str)))

for line in s1:
    print("{} -> max: {}, min: {}".format(line, max(line.split(',')), min(line.split(','))))

 '''
5,6,7,8,9 -> max: 9, min: 5
11,12,13,14 -> max: 14, min:  11 
17,18,19,20,21,22,23,24,25,26,27 -> max: 27, min: 17        
41,42 -> max: 42, min: 41                                     
44,45,46,47 -> max: 47, min: 44                             
'''

#3

After you get the s1

获得s1后

s2=s1.str.split(',',expand=True).apply(pd.to_numeric)
s2.max(1)
Out[29]: 
0     9.0
1    14.0
2    27.0
3    42.0
4    47.0
dtype: float64
s2.min(1)
Out[30]: 
0     5.0
1    11.0
2    17.0
3    41.0
4    44.0
dtype: float64

If you like int , you can adding astype(int) at the end

如果你喜欢int,你可以在最后添加astype(int)

#4

Here's what you can do with apply function

以下是使用apply函数可以执行的操作

min_max = s1.apply(lambda x: (min(map(int, x.split(','))), 
                              max(map(int, x.split(',')))))

for min_, max_ in min_max:
  print (min_, max_)

execution time:

In [10]: timeit s1.apply(lambda x: (min(map(int, x.split(','))), max(map(int, x.split(',')))))
109 µs ± 445 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

#1