使用 python 熊猫分箱列
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Binning column with python pandas
提问by Night Walker
I have a Data Frame column with numeric values:
我有一个带有数值的数据框列：
df['percentage'].head()
46.5
44.2
100.0
42.12
I want to see the column as bin counts:
我想将列视为 bin 计数：
bins = [0, 1, 5, 10, 25, 50, 100]
How can I get the result as bins with their value counts?
我怎样才能得到结果与他们的垃圾箱value counts？
[0, 1] bin amount
[1, 5] etc
[5, 10] etc
......
回答by jezrael
You can use pandas.cut:
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
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_countsor groupbyand aggregate size:
...然后value_counts或groupby和聚合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 cutreturn categorical.
默认cut返回categorical。
Seriesmethods like Series.value_counts()will use all categories, even if some categories are not present in the data, operations in categorical.
Series方法 likeSeries.value_counts()将使用所有类别，即使数据中不存在某些类别， categorical 中的操作。
回答by Erfan
Using numbamodule for speed up.
使用numba模块加速。
On big datasets (500k >) pd.cutcan be quite slow for binning data.
在大数据集上 ( 500k >)pd.cut对数据进行分箱可能会很慢。
I wrote my own function in numbawith just in time compilation, which is roughly 16xfaster:
我numba用即时编译编写了自己的函数，这大致16x更快：
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:
可选：您还可以将其作为字符串映射到 bin：
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:
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)
