Access a DataFrame element with Pandas, when we don't use the index-CodePudding

When we use the index to select a specific row, we can access an element with .loc, .iloc, etc.:

df = pd.DataFrame([[1, 2], [3, 4]], columns=["col1", "col2"], index=["aa", "bb"])
x = df.loc["aa", "col2"]
print(x, type(x))  # 2 <class 'numpy.int64'>

But when our id column is not the index, such as:

   id  col1  col2
0  aa     1     2
1  bb     3     4

what is the natural Pandas way to access the element of column col2 and of row of id equal to aa?

This doesn't work:

df = pd.DataFrame([["aa", 1, 2], ["bb", 3, 4]], columns=["id", "col1", "col2"])
x = df[df["id"] == "aa"]["col2"]
print(x, type(x))
# 0    2
# Name: col2, dtype: int64 <class 'pandas.core.series.Series'>

because it outputs a Series and not a number as expected. Is there a more standard way than adding an extra [0]:

x = df[df["id"] == "aa"]["col2"][0]  # 2, as expected

TL;DR: Why is df[df["id"] == "aa"]["col2"] a Series and not an element?

CodePudding user response：

If you need to return a scalar after applying a boolean mask, you can use pandas.Series.values :

x = df.loc[df['id'] == 'aa', 'col2'].values[0]

print(x, type(x))
2 <class 'numpy.int64'> #output

CodePudding user response：

what is the natural Pandas way to access the element of column col2 and of row of id equal to aa?

Unfortunately not exist, need convert id to index first for avoid one element Series, so need select by position [0].