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我喜欢像时钟一样循环旋转行,但我希望每一行都会根据“n_roll”的列进行不同的旋转

所以如果我有那个df

data={"col1":[2,3,4,5],
"col2":[4,2,4,6],
"col3":[7,6,9,11],
"col4":[14,11,22,8],
"name":["A","A","V","A"],
"n_roll":[1,2,2,3]}
df=pd.DataFrame.from_dict(data)
df

所以我希望它看起来像这样

data={"col1":[14,6,9,6],
"col2":[2,11,22,11],
"col3":[4,3,4,8],
"col4":[7,2,4,5],
"name":["A","A","V","A"],
"n_roll":[1,2,2,3]}
df=pd.DataFrame.from_dict(data)
df

也许是这样的: coll_to_roll=["col1","col2","col3","col4"]

df[coll_to_roll] = np.roll(df[coll_to_roll],1,df["n_roll"])
4

1 回答 1

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您可以通过将 DataFrame 和 column 转换为 numpy 数组来重用现有函数:

coll_to_roll=["col1","col2","col3","col4"]

from skimage.util.shape import view_as_windows as viewW

#https://stackoverflow.com/a/51613442
def strided_indexing_roll(a, r):
    # Concatenate with sliced to cover all rolls
    a_ext = np.concatenate((a,a[:,:-1]),axis=1)

    # Get sliding windows; use advanced-indexing to select appropriate ones
    n = a.shape[1]
    return viewW(a_ext,(1,n))[np.arange(len(r)), (n-r)%n,0]


df[coll_to_roll]=strided_indexing_roll(df[coll_to_roll].to_numpy(),df["n_roll"].to_numpy())
print (df)
   col1  col2  col3  col4 name  n_roll
0    14     2     4     7    A       1
1     6    11     3     2    A       2
2     9    22     4     4    V       2
3     6    11     8     5    A       3
于 2020-09-22T10:33:16.217 回答