标签归档:binning

用python熊猫装箱列

问题:用python熊猫装箱列

我有一个带有数值的数据框列:

df['percentage'].head()
46.5
44.2
100.0
42.12

我想查看该列作为箱数:

bins = [0, 1, 5, 10, 25, 50, 100]

我如何将结果作为垃圾箱value counts

[0, 1] bin amount
[1, 5] etc 
[5, 10] etc 
......

I have a Data Frame column with numeric values:

df['percentage'].head()
46.5
44.2
100.0
42.12

I want to see the column as bin counts:

bins = [0, 1, 5, 10, 25, 50, 100]

How can I get the result as bins with their value counts?

[0, 1] bin amount
[1, 5] etc 
[5, 10] etc 
......

回答 0

您可以使用pandas.cut

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = pd.cut(df['percentage'], bins)
print (df)
   percentage     binned
0       46.50   (25, 50]
1       44.20   (25, 50]
2      100.00  (50, 100]
3       42.12   (25, 50]

bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
df['binned'] = pd.cut(df['percentage'], bins=bins, labels=labels)
print (df)
   percentage binned
0       46.50      5
1       44.20      5
2      100.00      6
3       42.12      5

numpy.searchsorted

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = np.searchsorted(bins, df['percentage'].values)
print (df)
   percentage  binned
0       46.50       5
1       44.20       5
2      100.00       6
3       42.12       5

…然后value_countsor groupby和合计size

s = pd.cut(df['percentage'], bins=bins).value_counts()
print (s)
(25, 50]     3
(50, 100]    1
(10, 25]     0
(5, 10]      0
(1, 5]       0
(0, 1]       0
Name: percentage, dtype: int64

s = df.groupby(pd.cut(df['percentage'], bins=bins)).size()
print (s)
percentage
(0, 1]       0
(1, 5]       0
(5, 10]      0
(10, 25]     0
(25, 50]     3
(50, 100]    1
dtype: int64

默认cut返回categorical

Series像这样的方法Series.value_counts()将使用所有类别,即使数据中不存在某些类别,也可以使用categorical 操作

You can use pandas.cut:

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = pd.cut(df['percentage'], bins)
print (df)
   percentage     binned
0       46.50   (25, 50]
1       44.20   (25, 50]
2      100.00  (50, 100]
3       42.12   (25, 50]

bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
df['binned'] = pd.cut(df['percentage'], bins=bins, labels=labels)
print (df)
   percentage binned
0       46.50      5
1       44.20      5
2      100.00      6
3       42.12      5

Or numpy.searchsorted:

bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = np.searchsorted(bins, df['percentage'].values)
print (df)
   percentage  binned
0       46.50       5
1       44.20       5
2      100.00       6
3       42.12       5

…and then value_counts or groupby and aggregate size:

s = pd.cut(df['percentage'], bins=bins).value_counts()
print (s)
(25, 50]     3
(50, 100]    1
(10, 25]     0
(5, 10]      0
(1, 5]       0
(0, 1]       0
Name: percentage, dtype: int64

s = df.groupby(pd.cut(df['percentage'], bins=bins)).size()
print (s)
percentage
(0, 1]       0
(1, 5]       0
(5, 10]      0
(10, 25]     0
(25, 50]     3
(50, 100]    1
dtype: int64

By default cut return categorical.

Series methods like Series.value_counts() will use all categories, even if some categories are not present in the data, operations in categorical.


回答 1

使用numba模块加速。

在大型数据集(500k >pd.cut上,对数据进行合并可能会非常慢。

我编写了自己的函数,numba并进行了及时编译,这大约16x要快一些:

from numba import njit

@njit
def cut(arr):
    bins = np.empty(arr.shape[0])
    for idx, x in enumerate(arr):
        if (x >= 0) & (x < 1):
            bins[idx] = 1
        elif (x >= 1) & (x < 5):
            bins[idx] = 2
        elif (x >= 5) & (x < 10):
            bins[idx] = 3
        elif (x >= 10) & (x < 25):
            bins[idx] = 4
        elif (x >= 25) & (x < 50):
            bins[idx] = 5
        elif (x >= 50) & (x < 100):
            bins[idx] = 6
        else:
            bins[idx] = 7

    return bins
cut(df['percentage'].to_numpy())

# array([5., 5., 7., 5.])

可选:您还可以将其作为字符串映射到垃圾箱:

a = cut(df['percentage'].to_numpy())

conversion_dict = {1: 'bin1',
                   2: 'bin2',
                   3: 'bin3',
                   4: 'bin4',
                   5: 'bin5',
                   6: 'bin6',
                   7: 'bin7'}

bins = list(map(conversion_dict.get, a))

# ['bin5', 'bin5', 'bin7', 'bin5']

速度比较

# create dataframe of 8 million rows for testing
dfbig = pd.concat([df]*2000000, ignore_index=True)

dfbig.shape

# (8000000, 1)
%%timeit
cut(dfbig['percentage'].to_numpy())

# 38 ms ± 616 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
pd.cut(dfbig['percentage'], bins=bins, labels=labels)

# 215 ms ± 9.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Using numba module for speed up.

On big datasets (500k >) pd.cut can be quite slow for binning data.

I wrote my own function in numba with just in time compilation, which is roughly 16x faster:

from numba import njit

@njit
def cut(arr):
    bins = np.empty(arr.shape[0])
    for idx, x in enumerate(arr):
        if (x >= 0) & (x < 1):
            bins[idx] = 1
        elif (x >= 1) & (x < 5):
            bins[idx] = 2
        elif (x >= 5) & (x < 10):
            bins[idx] = 3
        elif (x >= 10) & (x < 25):
            bins[idx] = 4
        elif (x >= 25) & (x < 50):
            bins[idx] = 5
        elif (x >= 50) & (x < 100):
            bins[idx] = 6
        else:
            bins[idx] = 7

    return bins
cut(df['percentage'].to_numpy())

# array([5., 5., 7., 5.])

Optional: you can also map it to bins as strings:

a = cut(df['percentage'].to_numpy())

conversion_dict = {1: 'bin1',
                   2: 'bin2',
                   3: 'bin3',
                   4: 'bin4',
                   5: 'bin5',
                   6: 'bin6',
                   7: 'bin7'}

bins = list(map(conversion_dict.get, a))

# ['bin5', 'bin5', 'bin7', 'bin5']

Speed comparison:

# create dataframe of 8 million rows for testing
dfbig = pd.concat([df]*2000000, ignore_index=True)

dfbig.shape

# (8000000, 1)
%%timeit
cut(dfbig['percentage'].to_numpy())

# 38 ms ± 616 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
pd.cut(dfbig['percentage'], bins=bins, labels=labels)

# 215 ms ± 9.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)