Get first row of dataframe in Python Pandas based on criteria

Question

Let\'s say that I have a dataframe like this one

import pandas as pd
df = pd.DataFrame([[1, 2, 1], [1, 3, 2], [4, 6, 3], [4, 3, 4], [5, 4, 5]], columns=[\'A\         


        

Answer 1

This tutorial is a very good one for pandas slicing. Make sure you check it out. Onto some snippets... To slice a dataframe with a condition, you use this format:

>>> df[condition]

This will return a slice of your dataframe which you can index using iloc. Here are your examples:

1. Get first row where A > 3 (returns row 2)

   >>> df[df.A > 3].iloc[0]
   A    4
   B    6
   C    3
   Name: 2, dtype: int64
   

If what you actually want is the row number, rather than using iloc, it would be df[df.A > 3].index[0].

2. Get first row where A > 4 AND B > 3:

   >>> df[(df.A > 4) & (df.B > 3)].iloc[0]
   A    5
   B    4
   C    5
   Name: 4, dtype: int64
   

3. Get first row where A > 3 AND (B > 3 OR C > 2) (returns row 2)

   >>> df[(df.A > 3) & ((df.B > 3) | (df.C > 2))].iloc[0]
   A    4
   B    6
   C    3
   Name: 2, dtype: int64
   

Now, with your last case we can write a function that handles the default case of returning the descending-sorted frame:

>>> def series_or_default(X, condition, default_col, ascending=False):
...     sliced = X[condition]
...     if sliced.shape[0] == 0:
...         return X.sort_values(default_col, ascending=ascending).iloc[0]
...     return sliced.iloc[0]
>>> 
>>> series_or_default(df, df.A > 6, 'A')
A    5
B    4
C    5
Name: 4, dtype: int64

As expected, it returns row 4.

Answer 2

For existing matches, use query:

df.query(' A > 3' ).head(1)
Out[33]: 
   A  B  C
2  4  6  3

df.query(' A > 4 and B > 3' ).head(1)
Out[34]: 
   A  B  C
4  5  4  5

df.query(' A > 3 and (B > 3 or C > 2)' ).head(1)
Out[35]: 
   A  B  C
2  4  6  3

Answer 3

you can take care of the first 3 items with slicing and head:

1. df[df.A>=4].head(1)
2. df[(df.A>=4)&(df.B>=3)].head(1)
3. df[(df.A>=4)&((df.B>=3) * (df.C>=2))].head(1)

The condition in case nothing comes back you can handle with a try or an if...

try:
    output = df[df.A>=6].head(1)
    assert len(output) == 1
except: 
    output = df.sort_values('A',ascending=False).head(1)

Answer 4

For the point that 'returns the value as soon as you find the first row/record that meets the requirements and NOT iterating other rows', the following code would work:

def pd_iter_func(df):
    for row in df.itertuples():
        # Define your criteria here
        if row.A > 4 and row.B > 3:
            return row

It is more efficient than Boolean Indexing when it comes to a large dataframe.

To make the function above more applicable, one can implements lambda functions:

def pd_iter_func(df: DataFrame, criteria: Callable[[NamedTuple], bool]) -> Optional[NamedTuple]:
    for row in df.itertuples():
        if criteria(row):
            return row

pd_iter_func(df, lambda row: row.A > 4 and row.B > 3)

---

As mentioned in the answer to the 'mirror' question, pandas.Series.idxmax would also be a nice choice.

def pd_idxmax_func(df, mask):
    return df.loc[mask.idxmax()]

pd_idxmax_func(df, (df.A > 4) & (df.B > 3))