如何检查 pandas dataframe 中的列之间的冲突？

Question

I'm working on a Dataframe which contains multiple possible values from three different sources for a single item, which is in the index, such as:

import pandas as pd
import numpy as np

inp = [
    {"Item": "Item1", "Local A": np.nan, "Local B": 6, "Local C": 5},
    {"Item": "Item2", "Local A": 6, "Local B": 7, "Local C": 5},
    {"Item": "Item3", "Local A": np.nan, "Local B": np.nan, "Local C": 5},
    {"Item": "Item4", "Local A": 5, "Local B": 5, "Local C": 5},
    {"Item": "Item5", "Local A": 5, "Local B": np.nan, "Local C": 5},
]
df = pd.DataFrame(inp)
print(df)

Output:

    Item  Local A  Local B  Local C
0  Item1      NaN      6.0        5
1  Item2      6.0      7.0        5
2  Item3      NaN      NaN        5
3  Item4      5.0      5.0        5
4  Item5      5.0      NaN        5

My goal is to create a column which specifies if there is conflict between sources when there are multiple non-null values for an index (some cells are empty).

Ideal Output:

    Item  Local A  Local B  Local C Conflict
0  Item1      NaN      6.0        5      yes
1  Item2      6.0      7.0        5      yes
2  Item3      NaN      NaN        5      NaN
3  Item4      5.0      5.0        5      NaN
4  Item5      5.0      NaN        5      NaN

In order to do that I decided to build a filter that checks if the three sources are non-null and if they are different.

I built the filters for the three other cases consisting of two values being available for an index.

condition1 = (
    df["Local A"].notnull() & df["Local B"].notnull() & df["Local C"].notnull()
) & ~(df["Local A"] == df["Local B"] == df["Local C"])

condition2 = (df["Local A"].notnull() & df["Local B"].notnull()) & ~(
    df["Local A"] == df["Local B"]
)

condition3 = (df["Local B"].notnull() & df["Local C"].notnull()) & ~(
    df["Local B"] == df["Local C"]
)

condition4 = (df["Local A"].notnull() & df["Local C"].notnull()) & ~(
    df["Local A"] == df["Local C"]
)


df.loc[condition1 | condition2 | condition3 | condition4, "Conflict"] = "yes"

This solution of enumerating the different possible outcomes is not very elegant but I wasn't able to find a simpler alternative. Moreover, I get the following error while running the script:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I've seen this a few times and was able to find the cause, but I just can't figure this one out. It seems that I'm comparing Bool series instead of individual cases like I want to.

Answer 1

IIUC, try:

df['Conflict'] = np.where((df.iloc[:, 1:].nunique(axis=1) != 1),'Yes',np.nan)

Output:

    Item  Local A  Local B  Local C Conflict
0  Item1      NaN      6.0        5      Yes
1  Item2      6.0      7.0        5      Yes
2  Item3      NaN      NaN        5      nan
3  Item4      5.0      5.0        5      nan
4  Item5      5.0      NaN        5      nan