pandas 使用 np.where 但如果条件为 False 则保持现有值
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Using np.where but maintaining exisitng values if condition is False
提问by Woody Pride
I like np.where, but have never fully got to grip with it.
我喜欢 np.where，但从来没有完全掌握它。
I have a dataframe lets say it looks like this:
我有一个数据框可以说它看起来像这样：
import pandas as pd
import numpy as np
from numpy import nan as NA
DF = pd.DataFrame({'a' : [ 3, 0, 1, 0, 1, 14, 2, 0, 0, 0, 0],
'b' : [ 3, 0, 1, 0, 1, 14, 2, 0, 0, 0, 0],
'c' : [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
'd' : [5, 1, 2 ,1, 1 ,22, 30, 1, 0, 0, 0]})
Now what I want to do is replace the 0 values with NaN values when all row values are zero. Critically I want to maintain whatever other values are in the row in the cases where all row values are not zero.
现在我想要做的是当所有行值都为零时用 NaN 值替换 0 值。至关重要的是，在所有行值都不为零的情况下，我想保留行中的任何其他值。
I want to do something like this:
我想做这样的事情：
cols = ['a', 'b', 'c', 'd']
condition = (DF[cols] == 0).all(axis=1)
for col in cols:
DF[col] = np.where(condition, NA, ???)
I put the ??? to indicate that I do not know what value to place there if the condition is False, I just want to preserve whatever is there already. Is this possible with np.where, or should I use another technique?
我把？？？为了表明如果条件为 False，我不知道在那里放置什么值，我只想保留已经存在的任何值。这可以通过 np.where 实现，还是应该使用其他技术？
回答by JaminSore
There is a pandas.Seriesmethod (whereincidentally) for exactly this kind of task. It seems a little backward at first, but from the documentation.
对于这种任务，有一种pandas.Series方法（where顺便说一下）。起初似乎有点落后，但从文档来看。
Series.where(cond, other=nan, inplace=False, axis=None, level=None, try_cast=False, raise_on_error=True)
Return an object of same shape as self and whose corresponding entries are from self where cond is True and otherwise are from other.
Series.where(cond, other=nan, inplace=False, axis=None, level=None, try_cast=False, raise_on_error=True)
返回一个与 self 形状相同的对象，其对应的条目来自 self ，其中 cond 为 True ，否则来自 other 。
So, your example would become
所以，你的例子会变成
cols = ['a', 'b', 'c', 'd']
condition = (DF[cols] == 0).all(axis=1)
for col in cols:
DF[col].where(~condition, np.nan, inplace=True)
But, if all you're trying to do is replace rows of all zeros for specific set of columns with NA, you could do this instead
但是，如果您要做的只是将特定列集的全零行替换为NA，则可以改为执行此操作
DF.loc[condition, cols] = NA
EDIT
编辑
To answer your original question, np.wherefollows the same broadcasting rulesas other array operations so you would replace ???with DF[col], changing your example to:
要回答您的原始问题，请np.where遵循与其他数组操作相同的广播规则，因此您将替换???为DF[col]，将您的示例更改为：
cols = ['a', 'b', 'c', 'd']
condition = (DF[cols] == 0).all(axis=1)
for col in cols:
DF[col] = np.where(condition, NA, DF[col])
回答by Arvind Subramaniam
You can do something like this:
你可以这样做：
array_binary = np.where(array[i]<threshold,0,1)
array_sparse = np.multiply(array_binary,np.ones_like(array))
do an element-wise multiplication of the binary array and an array of ones using np.multiply. Hence, the non-zero elements will be recovered/maintained. array_sparse is the sparse version of array
使用 np.multiply 对二进制数组和一个数组进行逐元素乘法。因此，非零元素将被恢复/维护。array_sparse 是数组的稀疏版本
