pandas 减去数据帧熊猫时的NaN
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NaNs when subtracting dataframes pandas
提问by Marcus H?gen? Bohman
I have two dataframes with only somewhat overlapping indices and columns.
我有两个数据框，它们的索引和列只有一些重叠。
old = pd.DataFrame(index = ['A', 'B', 'C'],
columns = ['k', 'l', 'm'],
data = abs(np.floor(np.random.rand(3, 3)*10)))
new = pd.DataFrame(index = ['A', 'B', 'C', 'D'],
columns = ['k', 'l', 'm', 'n'],
data = abs(np.floor(np.random.rand(4, 4)*10)))
I want to calculate the difference between them and tried
我想计算它们之间的差异并尝试
delta = new - old
This gives lots of NaNs where indices and columns do not match. I would like to treat the abscence of the indices and columns as zeroes, (old['n', 'D'] = 0). old will always be a subspace of new.
这给出了许多索引和列不匹配的 NaN。我想将索引和列的缺失视为零，(old['n', 'D'] = 0)。old 永远是 new 的子空间。
Any ideas?
有任何想法吗？
EDIT: I guess I didn't explain it thoroughly enough. I don't want to fill the delta dataframe with zeroes. I want to treat missing indices and columns in old as if they were zeroes. I would then get the value in new['n', 'D'] in delta instead of a NaN.
编辑：我想我没有足够彻底地解释它。我不想用零填充增量数据帧。我想将旧的缺失索引和列视为零。然后我会在 new['n', 'D'] 中获得 delta 而不是 NaN 中的值。
采纳答案by EdChum
Use subwith fill_value=0:
使用sub有fill_value=0：
In [15]:
old = pd.DataFrame(index = ['A', 'B', 'C'],
columns = ['k', 'l', 'm'],
data = abs(np.floor(np.random.rand(3, 3)*10)))
?
new = pd.DataFrame(index = ['A', 'B', 'C', 'D'],
columns = ['k', 'l', 'm', 'n'],
data = abs(np.floor(np.random.rand(4, 4)*10)))
delta = new.sub(old, fill_value=0)
delta
Out[15]:
k l m n
A 0 3 -9 7
B 0 -2 1 8
C -4 1 1 7
D 8 6 0 6
