如何循环滚动/平移熊猫系列

Question

我喜欢像循环时钟一样旋转行，但我喜欢每一行将根据"；n_ roll"；

所以如果我有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个

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"])

Answer 1

您可以通过将DataFrame和列转换为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