我如何有效地获得一个NumPy数组中每个唯一值的频率计数?
>>> x = np.array([1,1,1,2,2,2,5,25,1,1])
>>> freq_count(x)
[(1, 5), (2, 3), (5, 1), (25, 1)]
我如何有效地获得一个NumPy数组中每个唯一值的频率计数?
>>> x = np.array([1,1,1,2,2,2,5,25,1,1])
>>> freq_count(x)
[(1, 5), (2, 3), (5, 1), (25, 1)]
当前回答
为了计算唯一的非整数——类似于Eelco Hoogendoorn的答案,但速度要快得多(在我的机器上是5倍),我使用了weave。内联组合numpy。只有一点c代码;
import numpy as np
from scipy import weave
def count_unique(datain):
"""
Similar to numpy.unique function for returning unique members of
data, but also returns their counts
"""
data = np.sort(datain)
uniq = np.unique(data)
nums = np.zeros(uniq.shape, dtype='int')
code="""
int i,count,j;
j=0;
count=0;
for(i=1; i<Ndata[0]; i++){
count++;
if(data(i) > data(i-1)){
nums(j) = count;
count = 0;
j++;
}
}
// Handle last value
nums(j) = count+1;
"""
weave.inline(code,
['data', 'nums'],
extra_compile_args=['-O2'],
type_converters=weave.converters.blitz)
return uniq, nums
配置文件信息
> %timeit count_unique(data)
> 10000 loops, best of 3: 55.1 µs per loop
Eelco的纯numpy版本:
> %timeit unique_count(data)
> 1000 loops, best of 3: 284 µs per loop
Note
这里存在冗余(unique也执行排序),这意味着可以通过将唯一功能放入c-code循环中来进一步优化代码。
其他回答
为了计算唯一的非整数——类似于Eelco Hoogendoorn的答案,但速度要快得多(在我的机器上是5倍),我使用了weave。内联组合numpy。只有一点c代码;
import numpy as np
from scipy import weave
def count_unique(datain):
"""
Similar to numpy.unique function for returning unique members of
data, but also returns their counts
"""
data = np.sort(datain)
uniq = np.unique(data)
nums = np.zeros(uniq.shape, dtype='int')
code="""
int i,count,j;
j=0;
count=0;
for(i=1; i<Ndata[0]; i++){
count++;
if(data(i) > data(i-1)){
nums(j) = count;
count = 0;
j++;
}
}
// Handle last value
nums(j) = count+1;
"""
weave.inline(code,
['data', 'nums'],
extra_compile_args=['-O2'],
type_converters=weave.converters.blitz)
return uniq, nums
配置文件信息
> %timeit count_unique(data)
> 10000 loops, best of 3: 55.1 µs per loop
Eelco的纯numpy版本:
> %timeit unique_count(data)
> 1000 loops, best of 3: 284 µs per loop
Note
这里存在冗余(unique也执行排序),这意味着可以通过将唯一功能放入c-code循环中来进一步优化代码。
老问题,但我想提供我自己的解决方案,这是最快的,使用普通列表而不是np。数组作为输入(或首先转移到列表),基于我的台架测试。
如果你也遇到这种情况,请检查一下。
def count(a):
results = {}
for x in a:
if x not in results:
results[x] = 1
else:
results[x] += 1
return results
例如,
>>>timeit count([1,1,1,2,2,2,5,25,1,1]) would return:
100000个循环,最好的3:2.26µs每循环
>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]))
100000个回路,最好的3:8.8µs每回路
>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]).tolist())
100000个回路,最佳3:5.85µs每回路
而公认的答案会更慢,而scipy.stats.itemfreq解决方案更糟糕。
更深入的测试并没有证实所制定的期望。
from zmq import Stopwatch
aZmqSTOPWATCH = Stopwatch()
aDataSETasARRAY = ( 100 * abs( np.random.randn( 150000 ) ) ).astype( np.int )
aDataSETasLIST = aDataSETasARRAY.tolist()
import numba
@numba.jit
def numba_bincount( anObject ):
np.bincount( anObject )
return
aZmqSTOPWATCH.start();np.bincount( aDataSETasARRAY );aZmqSTOPWATCH.stop()
14328L
aZmqSTOPWATCH.start();numba_bincount( aDataSETasARRAY );aZmqSTOPWATCH.stop()
592L
aZmqSTOPWATCH.start();count( aDataSETasLIST );aZmqSTOPWATCH.stop()
148609L
参考下面关于影响小型数据集大量重复测试结果的缓存和其他ram内副作用的评论。
使用pandas模块:
>>> import pandas as pd
>>> import numpy as np
>>> x = np.array([1,1,1,2,2,2,5,25,1,1])
>>> pd.value_counts(x)
1 5
2 3
25 1
5 1
dtype: int64
import pandas as pd
import numpy as np
x = np.array( [1,1,1,2,2,2,5,25,1,1] )
print(dict(pd.Series(x).value_counts()))
这会给你: {1,5, 2,3, 5:1, 25: 1}
from collections import Counter
x = array( [1,1,1,2,2,2,5,25,1,1] )
mode = counter.most_common(1)[0][0]