python – 基于pandas数据帧替换numpy 2D数组中的值

2023年2月5日 173次阅读 
>>> arr
array([[ 0., 10.,  0., ...,  0.,  0.,  0.],
           [ 0.,  4.,  0., ...,  6.,  0.,  9.],
           [ 0.,  0.,  0., ...,  0.,  0.,  0.],
           ..., 
           [ 0.,  0.,  0., ...,  0.,  0.,  0.],
           [ 0.,  0.,  0., ...,  2.,  0.,  0.],
           [ 0.,  0.,  0., ...,  0.,  3.,  0.]])

在上面的numpy数组中,我想将与dataf(df_A)中的country_codes列匹配的每个值替换为df_A中的continent_codes列中的值. df_A看起来像：

  country_codes   continent_codes
0              4      4
1              8      3
2             12      5
3             16      6
4             24      5

现在,我遍历数据框并使用numpy索引表示法替换.鉴于iterrows()往往很慢,是否有更直接/矢量化的方法来做到这一点？

for index, row in self.df_A.iterrows():
    arr[arr == row['country_codes']] = row['continent_codes']

最佳答案 方法#1：使用
np.searchsorted和
np.in1d的一种矢量化方法如下所示 –

# Store country_codes and continent_codes column data for further usage
oldval = np.array(df['country_codes'])
newval = np.array(df['continent_codes'])

# Mask of elements to be changed
mask = np.in1d(arr,oldval)

# Indices for each match from oldval in arr
idx = np.searchsorted(oldval,arr.ravel()[mask])

# Using the mask put selective elements from continent_codes column into arr
arr.ravel()[mask] = newval[idx]

样品运行 –

>>> arr   # Original 2D array
array([[23,  4, 23,  5,  8],
       [ 3,  6,  8,  5, 11],
       [16, 24, 15,  4, 10],
       [ 4, 16, 10,  8,  1]])
>>> df
   country_codes  continent_codes
0              4                4
1              8                3
2             12                5
3             16                6
4             24                5

>>> oldval = np.array(df['country_codes'])
>>> newval = np.array(df['continent_codes'])
>>> mask = np.in1d(arr,oldval)
>>> idx = np.searchsorted(oldval,arr.ravel()[mask])
>>> arr.ravel()[mask] = newval[idx]

>>> mask.reshape(arr.shape)  # Mask array depiciting which elements were updated
array([[False,  True, False, False,  True],
       [False, False,  True, False, False],
       [ True,  True, False,  True, False],
       [ True,  True, False,  True, False]], dtype=bool)
>>> arr  # Updated 2D array
array([[23,  4, 23,  5,  3],
       [ 3,  6,  3,  5, 11],
       [ 6,  5, 15,  4, 10],
       [ 4,  6, 10,  3,  1]])

方法#2：作为变体,您还可以创建掩码,并在np.searchsorted(oldval,arr,’left’)和np.searchsorted(oldval,arr,’right’)之间进行比较,如this question解决方案中所述稍后再次使用np.searchsorted(oldval,arr,’left’),同时将值放入arr以获得更有效的解决方案,如下所示 –

# Store country_codes and continent_codes column data for further usage
oldval = np.array(df['country_codes'])
newval = np.array(df['continent_codes'])

# Left and right indices for each match from oldval in arr
left_idx = np.searchsorted(oldval,arr,'left')
right_idx = np.searchsorted(oldval,arr,'right')

# Mask of elements to be changed
mask = left_idx!=right_idx

# Using the mask put selective elements from continent_codes column into arr
arr[mask] = newval[left_idx[mask]]

运行时测试并验证输出

功能定义 –

def original_app(arr,df):
    for index, row in df.iterrows():
        arr[arr == row['country_codes']] = row['continent_codes']

def vectorized_app1(arr,df):
    oldval = np.array(df['country_codes'])
    newval = np.array(df['continent_codes'])
    mask = np.in1d(arr,oldval)
    idx = np.searchsorted(oldval,arr.ravel()[mask])
    arr.ravel()[mask] = newval[idx]

def vectorized_app2(arr,df):
    oldval = np.array(df['country_codes'])
    newval = np.array(df['continent_codes'])
    left_idx = np.searchsorted(oldval,arr,'left')
    right_idx = np.searchsorted(oldval,arr,'right')
    mask = left_idx!=right_idx
    arr[mask] = newval[left_idx[mask]]

验证输出 –

In [195]: # Input array
     ...: arr = np.random.randint(0,100000,(1000,1000))
     ...: 
     ...: # Setup input dataframe
     ...: N = 1000
     ...: oldvals = np.unique(np.random.randint(0,100000,N))
     ...: newvals = np.random.randint(0,9,(oldvals.size))
     ...: df=pd.DataFrame({'country_codes':oldvals,'continent_codes':newvals})
     ...: df = df.reindex_axis(sorted(df.columns)[::-1], axis=1)
     ...: 
     ...: # Make copies for input array for funcs to update them
     ...: arrc1 = arr.copy()
     ...: arrc2 = arr.copy()
     ...: arrc3 = arr.copy()
     ...: 

In [196]: # Verify outputs
     ...: original_app(arrc1,df)
     ...: vectorized_app1(arrc2,df)
     ...: vectorized_app2(arrc3,df)
     ...: 

In [197]: np.allclose(arrc1,arrc2)
Out[197]: True

In [198]: np.allclose(arrc1,arrc3)
Out[198]: True

计时 –

In [199]: # Make copies for input array for funcs to update them
     ...: arrc1 = arr.copy()
     ...: arrc2 = arr.copy()
     ...: arrc3 = arr.copy()
     ...: 

In [200]: %timeit original_app(arrc1,df)
1 loops, best of 3: 2.79 s per loop

In [201]: %timeit vectorized_app1(arrc2,df)
1 loops, best of 3: 360 ms per loop

In [202]: %timeit vectorized_app2(arrc3,df)
1 loops, best of 3: 213 ms per loop
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