从熊猫切割中分类垃圾箱

Question

使用 pandas cut 我可以通过提供边缘来定义 bin，pandas 会创建像 (a, b] 这样的 bin。

我的问题是如何对垃圾箱进行排序（从最低到最高）？

import numpy as np
import pandas as pd

y = pd.Series(np.random.randn(100))

x1 = pd.Series(np.sign(np.random.randn(100)))
x2 = pd.cut(pd.Series(np.random.randn(100)), bins = [-3, -0.5, 0, 0.5, 3])

model = pd.concat([y, x1, x2], axis = 1, keys = ['Y', 'X1', 'X2'])

我有一个中间结果，其中保留了 bin 的顺序

int_output = model.groupby(['X1', 'X2']).mean().unstack()
int_output.columns = int_output.columns.get_level_values(1)

X2    (-3, -0.5]  (-0.5, 0]  (0, 0.5]  (0.5, 3]
X1                                             
-1.0    0.101475  -0.344419 -0.482992 -0.015179
 1.0    0.249961   0.484757 -0.066383 -0.249414

但随后我会进行其他任意更改 bin 顺序的操作：

output = pd.concat(int_output.to_dict('series'), axis = 1)

      (-0.5, 0]  (-3, -0.5]  (0, 0.5]  (0.5, 3]
X1                                             
-1.0  -0.344419    0.101475 -0.482992 -0.015179
 1.0   0.484757    0.249961 -0.066383 -0.249414

现在我想在条形图中绘制数据，但我希望将 bin 从最低 (-3, -0.5] 到最高 (0.5, 3]) 排序。

我想我可以通过操作字符串，在“，”上使用拆分然后清理括号来实现这一点，但我想知道是否有更好的方法。

Answer 1

主要问题是丢失ordered CategoricalIndex。

np.random.seed(12456)
y = pd.Series(np.random.randn(100))
x1 = pd.Series(np.sign(np.random.randn(100)))
x2 = pd.cut(pd.Series(np.random.randn(100)), bins = [-3, -0.5, 0, 0.5, 3])

model = pd.concat([y, x1, x2], axis = 1, keys = ['Y', 'X1', 'X2'])
int_output = model.groupby(['X1', 'X2']).mean().unstack()
int_output.columns = int_output.columns.get_level_values(1)

print (int_output)
X2    (-3, -0.5]  (-0.5, 0]  (0, 0.5]  (0.5, 3]
X1                                             
-1.0    0.230060  -0.079266 -0.079834 -0.064455
 1.0   -0.451351   0.268688  0.020091 -0.280218

print (int_output.columns)
CategoricalIndex(['(-3, -0.5]', '(-0.5, 0]', '(0, 0.5]', '(0.5, 3]'], 
                 categories=['(-3, -0.5]', '(-0.5, 0]', '(0, 0.5]', '(0.5, 3]'], 
                 ordered=True, name='X2', dtype='category')

output = pd.concat(int_output.to_dict('series'), axis = 1)
print (output)
      (-0.5, 0]  (-3, -0.5]  (0, 0.5]  (0.5, 3]
X1                                             
-1.0  -0.079266    0.230060 -0.079834 -0.064455
 1.0   0.268688   -0.451351  0.020091 -0.280218

print (output.columns)
Index(['(-0.5, 0]', '(-3, -0.5]', '(0, 0.5]', '(0.5, 3]'], dtype='object')

一个可能的解决方案是extract 来自output.columns 的第一个数字，创建助手系列并对其进行排序。最后reindex原创专栏：

cat = output.columns.str.extract('\((.*),', expand=False).astype(float)
a = pd.Series(cat, index=output.columns).sort_values()
print (a)
(-3, -0.5]   -3.0
(-0.5, 0]    -0.5
(0, 0.5]      0.0
(0.5, 3]      0.5
dtype: float64

output = output.reindex(columns=a.index)
print (output)
      (-3, -0.5]  (-0.5, 0]  (0, 0.5]  (0.5, 3]
X1                                             
-1.0    0.230060  -0.079266 -0.079834 -0.064455
 1.0   -0.451351   0.268688  0.020091 -0.280218

Answer 2

解决您上面强调的问题的一个简单方法是简单地重新排序列：

output[sorted(output.columns)]

Answer 3

我为此做了一个函数。

def dfsortbybins(df, col):
    """
    param df: pandas dataframe
    param col: name of column containing bins
    """
    d=dict(zip(bins,[float(s.split(',')[0].split('(')[1]) for s in bins]))
    df[f'{col} dfrankbybins']=df.apply(lambda x : d[x[col]] if not pd.isnull(x[col]) else x[col], axis=1)
    df=df.sort_values(f'{col} dfrankbybins').drop(f'{col} dfrankbybins',axis=1)
    return df

Answer 4

这是另一个功能。与其他解决方案不同，这在多种情况下对我有用。我想我会把它留在这里，希望它对将来遇到同样问题的人有用。

def sort_bins(bin_col):
    """
    Sorts bins after using pd.cut. Increasing order. Puts "NaN" bin at the beginning. 

    Input:
        bin_col: pd.series containing bins to be sorted

    """

    # Dictionary to store first value from each bin
    vals = {}

    # Iterate through all bins
    for i, item in enumerate(bin_col.unique()):

        # Check if bin is "nan", if yes, assign low value to put it at the beginning
        if item == "nan":
            vals[i] = -99999

        # If not "nan", get the first value from bin to sort later
        else:
            vals[i] = float(item.split(",")[0][1:])

    # Sort bins according to extracted first values
    ixs = list({k: v for k, v in \
                    sorted(vals.items(), key=lambda item: item[1])}.keys())

    # Make sorted list of bins
    sorted_bins = bin_col.unique()[list(ixs)]

    return sorted_bins

# Example, assuming "age_bin" column has the bins:
sorted_bins = sort_bins(df["age_bin"])