英文:
Mapping a pandas dataframe to a n-dimensional array, where each dimension corresponds to one of the x columns
问题
我有一个包含列 x0 x1 x2 x3 x4 和 y0 y1 y2 y3 y4 的数据帧。
前十行：
Id x0 x1 x2 x3 x4 y0 y1 y2 y3 y40 0 -5.0 -5.0 -5.0 -5.0 -5.0 268035854.2037072 0.94956508069182 3520.7568220782514 -412868933.038522 242572043.877278481 1 -5.0 -5.0 -5.0 -5.0 -4.5 268035883.40390667 0.94956508069182 3482.0382462663074 -412868933.038522 242572043.877278482 2 -5.0 -5.0 -5.0 -5.0 -4.0 268035901.1170006 0.94956508069182 3443.3196704543634 -412868933.038522 242572043.877278483 3 -5.0 -5.0 -5.0 -5.0 -3.5 268035911.8642905 0.94956508069182 3404.6010946424194 -412868933.038522 242572043.877278484 4 -5.0 -5.0 -5.0 -5.0 -3.0 268035918.38904288 0.94956508069182 3365.882518830476 -412868933.038522 242572043.877278485 5 -5.0 -5.0 -5.0 -5.0 -2.5 268035922.35671327 0.94956508069182 3327.163943018532 -412868933.038522 242572043.877278486 6 -5.0 -5.0 -5.0 -5.0 -2.0 268035924.7800574 0.94956508069182 3288.445367206588 -412868933.038522 242572043.877278487 7 -5.0 -5.0 -5.0 -5.0 -1.5 268035926.27763835 0.94956508069182 3249.726791394644 -412868933.038522 242572043.877278488 8 -5.0 -5.0 -5.0 -5.0 -1.0 268035927.2317166 0.94956508069182 3211.0082155827004 -412868933.038522 242572043.877278489 9 -5.0 -5.0 -5.0 -5.0 -0.5 268035927.8858225 0.94956508069182 3172.2896397707564 -412868933.038522 242572043.87727848
我执行了以下操作：
values = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].valuesvalues.shape
现在的形状是 (4084101, 5)。
我想要的形状是 (21, 21, 21, 21, 21, 5)（就好像我们有一个5D图，第一个维度是 x0，第二个是 x1，以此类推）。基本上，应该是 values[1, 0, 0, 0, 0] 来访问与 x0=-4.5、x1=-5、...、x4=-5 对应的元组 (y0, y1, y2, y3, y4)。
21 是因为 x0, ..., x4 的值范围从 -5 到 5，步长为 0.5，5 是因为 y0, y1, y2, y3, y4。
我尝试过 values = values.reshape(21, 21, 21, 21, 21, 5)，但当我执行 values[1][0][0][0][0] 时，我期望得到与 x1=-4.5, x2=-5, ..., x4=-5 相对应的值，但我没有得到。
我曾经有一个不太好的想法（从复杂性角度来看），那就是创建一个字典，其中键是元组 (x0, x1, x2, x3, x4)，属性是找到 y 值的索引。然后创建一个 np.zeros((21, 21, 21, 21, 21, 5)) 数据帧。
# 获取值values = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].values# 创建一个将 x0, x1, x2, x3, x4 值映射到索引的字典grid = {}for i, row in df_train.iterrows():x0, x1, x2, x3, x4 = [int((x + 5) / 0.5) for x in [row['x0'], row['x1'], row['x2'], row['x3'], row['x4']]]grid[(x0, x1, x2, x3, x4)] = i# 创建重塑后的数组reshaped_values = np.zeros((21, 21, 21, 21, 21, 5))for key, index in grid.items():reshaped_values[key[0]][key[1]][key[2]][key[3]][key[4]] = values[index]
但在我的计算机上运行几乎需要一分钟...看起来是一个糟糕的主意。
英文:
I have a dataframe with columns x0 x1 x2 x3 x4 and y0 y1 y2 y3 y4.
First ten rows:
Id x0 x1 x2 x3 x4 y0 y1 y2 y3 y40 0 -5.0 -5.0 -5.0 -5.0 -5.0 268035854.2037072 0.94956508069182 3520.7568220782514 -412868933.038522 242572043.877278481 1 -5.0 -5.0 -5.0 -5.0 -4.5 268035883.40390667 0.94956508069182 3482.0382462663074 -412868933.038522 242572043.877278482 2 -5.0 -5.0 -5.0 -5.0 -4.0 268035901.1170006 0.94956508069182 3443.3196704543634 -412868933.038522 242572043.877278483 3 -5.0 -5.0 -5.0 -5.0 -3.5 268035911.8642905 0.94956508069182 3404.6010946424194 -412868933.038522 242572043.877278484 4 -5.0 -5.0 -5.0 -5.0 -3.0 268035918.38904288 0.94956508069182 3365.882518830476 -412868933.038522 242572043.877278485 5 -5.0 -5.0 -5.0 -5.0 -2.5 268035922.35671327 0.94956508069182 3327.163943018532 -412868933.038522 242572043.877278486 6 -5.0 -5.0 -5.0 -5.0 -2.0 268035924.7800574 0.94956508069182 3288.445367206588 -412868933.038522 242572043.877278487 7 -5.0 -5.0 -5.0 -5.0 -1.5 268035926.27763835 0.94956508069182 3249.726791394644 -412868933.038522 242572043.877278488 8 -5.0 -5.0 -5.0 -5.0 -1.0 268035927.2317166 0.94956508069182 3211.0082155827004 -412868933.038522 242572043.877278489 9 -5.0 -5.0 -5.0 -5.0 -0.5 268035927.8858225 0.94956508069182 3172.2896397707564 -412868933.038522 242572043.87727848
I did this:
values = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].valuesvalues.shape
I now have shape (4084101, 5)
I would like to have shape (21, 21, 21, 21, 21, 5) (so that the first shape is x0, the second x1, like if we had a 5D graph). Basically, it should be values[1, 0, 0, 0, 0] to access the tuple (y0, y1, y2, y3, y4) corresponding to x0=-4.5, x1=-5, ..., x4=-5.
21 because values go from -5 to 5 for the x0, ..., x4 with step 0.5
and 5 because y0, y1, y2, y3, y4
I did values = values.reshape(21, 21, 21, 21, 21, 5)
But when I do values[1][0][0][0][0], I expected to have the value corresponding to x1=-4.5, x2=-5, ..., x4=-5 but I don't.
One bad idea that I had (complexity wise) was to make a dictionary in which keys are tuples (x0, x1, x2, x3, x4) and attributes the index where to find the y values.
And then fill a np.zeros((21, 21, 21, 21, 21, 5)) dataframe.
# Get the valuesvalues = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].values# Create a dictionary to map the x0, x1, x2, x3, x4 values to indicesgrid = {}for i, row in df_train.iterrows():x0, x1, x2, x3, x4 = [int((x + 5) / 0.5) for x in [row['x0'], row['x1'], row['x2'], row['x3'], row['x4']]]grid[(x0, x1, x2, x3, x4)] = i# Create the reshaped arrayreshaped_values = np.zeros((21, 21, 21, 21, 21, 5))for key, index in grid.items():reshaped_values[key[0]][key[1]][key[2]][key[3]][key[4]] = values[index]
but it takes almost a minute on my computer ... and looks like the worst idea ever.
答案1
得分: 2
你的代码有效，但我认为你的数据框没有排序。
df_train = df_train.sort_values(['x0', 'x1', 'x2', 'x3', 'x4'])values = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].values
英文:
Your code works but I think your dataframe is not sorted
df_train = df_train.sort_values(['x0', 'x1', 'x2', 'x3', 'x4'])values = df_train[['y0', 'y1', 'y2', 'y3', 'y4']].values
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