CODEWITHSUNDEEP
pythonpython-3.xpandasdataframelambda
Muds
Muds

Reputation: 4116

Inlie series of IFs present The truth value of a Series is ambiguous Error

I am trying to add and fill a column in a Pandas DataFrame based on a series of if conditions. My code looks like this

df['E'] = (('Text1;' if df['A'] else '') + ('Text2;' if df['B'] else '')) + ('Text3;' if df['C'] else '')

The dataframe looks like

A       B        C        D
TRUE    FALSE    FALSE    4
FALSE   TRUE     TRUE     7
FALSE   FALSE    FALSE    1

compiler errors out saying Truth Value being Ambiguous, I tried playing with parenthesis as suggested by SO but cant get this to work.

I have checked many other threads on SO around this error but none point to this problem.

Thanks

Upvotes: 1

Views: 76

Answers (1)

user17014529
user17014529

Reputation:

Your way not working because: if else with series and dfs are not good and there are many many QAs why it is wrong and what to do.

but still you could try np.select with 2^3 possibilities after checking them but you didn't show such attempt.

theres another way with matrix mul:

#pd.Index plays better than list when it comes to @ below
texts = pd.Index(["Text1;", "Text2;","Text3;"])
#choose interested columns
columns = ["A","B","C"]
#perform matrix mult and assign
df["E"] = df[columns] @ texts

matrix multiplication here with @ is "sum of products". It produces 3 such sum, one is for example "Text1;" * True + "Text2;" * False + "Text3;" * False for the first E result. this is usual matrix multiplcation as you might know

df becomes

       A      B      C  D             E
0   True  False  False  4        Text1;
1  False   True   True  7  Text2;Text3;
2  False  False  False  1

2^3 way with np.select. we form "logic table" from 000 to 111

conditions = [
    ~df.A & ~df.B & ~df.C,
    ~df.A & ~df.B &  df.C,
    ~df.A &  df.B & ~df.C,
    ~df.A &  df.B &  df.C,
     df.A & ~df.B & ~df.C,
     df.A & ~df.B &  df.C,
     df.A &  df.B & ~df.C,
     df.A &  df.B &  df.C,
]

chooses = [
    "",
    "Text3;",
    "Text2;",
    "Text2;Text3;",
    "Text1;",
    "Text1;Text3;",
    "Text1;Text2;",
    "Text1;Text2;Text3;",
]

df["E"] = np.select(conditions,chooses)

df is the same as above now.

Upvotes: 2

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