尝试在python中创建分组变量

Question

我有一列年龄值，我需要将其转换为18-29、30-39、40-49、50-59、60-69和70+以上的年龄范围：

对于df“文件”中某些数据的示例，我有：

并希望到达：

我尝试了以下方法：

file['agerange'] = file[['age']].apply(lambda x: "18-29" if (x[0] > 16
                                       or x[0] < 30) else "other")

我宁愿不只是进行分组，因为存储桶的大小也不统一，但如果可行的话，我会对此开放。

提前致谢！

Answer 1

嵌套循环不是这里最简单的解决方案吗？

import random
ages = [random.randint(18, 100) for _ in range(100)]
age_ranges = [(18,29), (30,39), (40,49), (50,59), (60,69),(70,)]

for a in ages:
        for r in age_ranges:
                if a >= r[0] and (len(r) == 1 or a < r[1]):
                        print a,r
                        break

Answer 2

您似乎正在使用Pandas库。 它们包括执行此操作的功能： http : //pandas.pydata.org/pandas-docs/version/0.16.0/genic/pandas.cut.html

这是我的尝试：

import pandas as pd

ages = pd.DataFrame([81, 42, 18, 55, 23, 35], columns=['age'])

bins = [18, 30, 40, 50, 60, 70, 120]
labels = ['18-29', '30-39', '40-49', '50-59', '60-69', '70+']
ages['agerange'] = pd.cut(ages.age, bins, labels = labels,include_lowest = True)

print(ages)

   age agerange
0   81      70+
1   42    40-49
2   18    18-29
3   55    50-59
4   23    18-29
5   35    30-39

Answer 3

您可以使用// 10作为键函数来使用itertools.groupby 。

In [10]: ages = [random.randint(18, 99) for _ in range(100)]

In [11]: [(key, list(group)) for (key, group) in itertools.groupby(sorted(ages), key=lambda x: x // 10)]
Out[11]: 
[(1, [18]),
 (2, [20, 21, 21, 22, 23, 24, 25, 26, 26, 26, 27, 27, 28]),
 (3, [30, 30, 32, 32, 34, 35, 36, 37, 37]),
 (4, [41, 42, 42, 43, 43, 44, 45, 47, 48]),
 (5, [50, 51, 52, 53, 54, 55, 56, 56, 56, 56, 57, 58, 58, 58, 58]),
 (6, [60, 61, 62, 62, 62, 65, 65, 66, 66, 66, 66, 67, 69, 69, 69]),
 (7, [71, 71, 72, 72, 73, 75, 75, 77, 77, 78]),
 (8, [83, 83, 83, 83, 84, 84, 85, 86, 86, 87, 87, 88, 89, 89, 89]),
 (9, [91, 91, 92, 92, 93, 94, 97, 97, 98, 98, 99, 99, 99])]

请记住， groupby需要排序的数据，因此请先排序。 或者使用字典和循环手动进行。

In [14]: groups = collections.defaultdict(list)

In [15]: for x in ages:
   ....:     groups[x//10].append(x)

In [16]: groups
Out[16]: defaultdict(<type 'list'>, {1: [18], 
             2: [26, 28, 21, 20, 26, 24, 21, 27, 25, 23, 27, 26, 22], 
             3: [37, 30, 32, 32, 35, 30, 36, 37, 34], 
             4: [45, 42, 43, 41, 47, 43, 48, 44, 42], 
             5: [52, 56, 58, 55, 58, 51, 58, 58, 57, 56, 53, 56, 50, 54, 56], 
             6: [69, 65, 62, 61, 65, 66, 66, 62, 69, 66, 67, 66, 60, 62, 69], 
             7: [71, 77, 71, 72, 77, 73, 78, 72, 75, 75], 
             8: [87, 83, 84, 86, 86, 83, 83, 87, 85, 83, 89, 88, 84, 89, 89], 
             9: [99, 92, 99, 98, 91, 94, 97, 92, 98, 97, 91, 93, 99]})

对于更复杂的分组，您可以使key功能任意复杂。 例如，为了将70岁及以上的所有人归为一组，请使用lambda x: min(x // 10, 7) 。 这适用于两种方法。 如果愿意，您甚至可以将密钥转换为字符串：

In [23]: keyfunc = lambda x: "{0}0-{0}9".format(x//10) if x < 70 else "70+"
In [24]: [(key, list(group)) for (key, group) in itertools.groupby(sorted(ages), key=keyfunc)]
Out[24]: 
[('10-19', [18]),
 ('20-29', [20, 21, 21, 22, 23, 24, 25, 26, 26, 26, 27, 27, 28]),
 ('30-39', [30, 30, 32, 32, 34, 35, 36, 37, 37]),
 ('40-49', [41, 42, 42, 43, 43, 44, 45, 47, 48]),
 ('50-59', [50, 51, 52, 53, 54, 55, 56, 56, 56, 56, 57, 58, 58, 58, 58]),
 ('60-69', [60, 61, 62, 62, 62, 65, 65, 66, 66, 66, 66, 67, 69, 69, 69]),
 ('70+',   [all the rest]]

Answer 4

一位朋友离线提出了以下有效的响应：def age_buckets（x）：如果x <30：返回'18 -29'elif x <40：返回'30 -39'elif x <50：返回'40 -49' elif x <60：返回'50 -59'elif x <70：返回'60 -69'elif x> = 70：返回'70 +'else：返回'other'

file['agerange'] = file.age.apply(age_buckets)

感谢所有对此努力的人！