pandas 一次迭代多行并重叠
问题描述
我有一个 pandas DataFrame,需要以 n 行块的形式输入下游函数(示例中为 print
).这些块可能有重叠的行.
I have a pandas DataFrame that need to be fed in chunks of n-rows into downstream functions (print
in the example). The chunks may have overlapping rows.
让我们从一个虚拟的 DataFrame 开始:
Let's start from a dummy DataFrame:
d = {'A':list(range(1000)), 'B':list(range(1000))}
df=pd.DataFrame(d)
对于具有 1 行重叠的 2 行块,我有以下代码:
In the case of a 2-rows chunks with 1-row overlap I have the following code:
a = df.index.values[:-1]
for i in a:
print(df.iloc[i:i+2])
输出是这样的:
...
A B
996 996 996
997 997 997
A B
997 997 997
998 998 998
A B
998 998 998
999 999 999
这正是我想要的.
是否有更好/更快的方法来迭代 pandas.DataFrame 的 n 行块?
Is there a better/faster approach to iterate over chunks of n-rows of a pandas.DataFrame?
解决方案
使用DataFrame.groupby
使用与 df
相同长度创建的辅助一维数组进行整数除法 - 索引值不重叠:
Use DataFrame.groupby
with integer division with helper 1d array created with same length like df
- index values are not overlapped:
d = {'A':list(range(5)), 'B':list(range(5))}
df=pd.DataFrame(d)
print (np.arange(len(df)) // 2)
[0 0 1 1 2]
for i, g in df.groupby(np.arange(len(df)) // 2):
print (g)
A B
0 0 0
1 1 1
A B
2 2 2
3 3 3
A B
4 4 4
对于重叠值进行编辑这个答案:
def chunker1(seq, size):
return (seq.iloc[pos:pos + size] for pos in range(0, len(seq)-1))
for i in chunker1(df,2):
print (i)
A B
0 0 0
1 1 1
A B
1 1 1
2 2 2
A B
2 2 2
3 3 3
A B
3 3 3
4 4 4
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