在不满足条件的所有行上过滤 Pandas 数据框
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filter a pandas data frame on all rows that do NOT meet a condition
提问by seth127
This seems simple, but I can't seem to figure it out. I know how to filter a pandas data frame to all rows that meet a condition, but when I want the opposite, I keep getting weird errors.
这看起来很简单，但我似乎无法弄清楚。我知道如何将 Pandas 数据框过滤到满足条件的所有行，但是当我想要相反的时候，我不断收到奇怪的错误。
Here is the example. (Context: a simple board game where pieces are on a grid and we're trying to give it a coordinate and return all adjacent pieces, but NOT the actual piece on that actual coordinate)
这是示例。（上下文：一个简单的棋盘游戏，棋子在网格上，我们试图给它一个坐标并返回所有相邻的棋子，但不是该实际坐标上的实际棋子）
import pandas as pd
import numpy as np
df = pd.DataFrame([[5,7, 'wolf'],
[5,6,'cow'],
[8, 2, 'rabbit'],
[5, 3, 'rabbit'],
[3, 2, 'cow'],
[7, 5, 'rabbit']],
columns = ['lat', 'long', 'type'])
coords = [5,7] #the coordinate I'm testing, a wolf
view = df[((coords[0] - 1) <= df['lat']) & (df['lat'] <= (coords[0] + 1)) \
& ((coords[1] - 1) <= df['long']) & (df['long'] <= (coords[1] + 1))]
view = view[not ((coords[0] == view['lat']) & (coords[1] == view['long'])) ]
print(view)
I thought the notshould just negate the boolean inside the parentheses that followed, but this doesn't seem to be how it works.
我认为not应该只是否定后面括号内的布尔值，但这似乎不是它的工作原理。
I want it to return the cow at 5,6 but NOT the wolf at 5,7 (because that's the current piece). Just to double check my logic, I did
我希望它在 5,6 处返回母牛，而不是在 5,7 处返回狼（因为这是当前的部分）。只是为了仔细检查我的逻辑，我做到了
me = view[(coords[0] == view['lat']) & (coords[1] == view['long'])]
print(me)
and this returned just the wolf, as I'd expected. So why can't I just put a notin front of that and get everything else? Or, more importantly, what do I do instead to get everything else.
正如我所料，这只是狼回来了。那么为什么我不能not在它前面放一个并获得其他所有东西呢？或者，更重要的是，我该怎么做才能获得其他一切。
回答by DeepSpace
As numpy(therefore pandas) use bitwise operators, you should replace notwith ~. This is also the reason you are using &and not and.
由于numpy（因此pandas）使用按位运算符，您应该替换not为~。这也是您使用&而不是的原因and。
import pandas as pd
df = pd.DataFrame({'a': [1, 2]})
print(df[~(df['a'] == 1)])
>> a
1 2
And using your example:
并使用您的示例：
import pandas as pd
import numpy as np
df = pd.DataFrame([[5,7, 'wolf'],
[5,6,'cow'],
[8, 2, 'rabbit'],
[5, 3, 'rabbit'],
[3, 2, 'cow'],
[7, 5, 'rabbit']],
columns = ['lat', 'long', 'type'])
coords = [5,7] #the coordinate I'm testing, a wolf
view = df[((coords[0] - 1) <= df['lat']) & (df['lat'] <= (coords[0] + 1)) \
& ((coords[1] - 1) <= df['long']) & (df['long'] <= (coords[1] + 1))]
view = view[~ ((coords[0] == view['lat']) & (coords[1] == view['long'])) ]
print(view)
>> lat long type
1 5 6 cow
