Question

目前，我正在使用两个lambda函数向数据集中的两列添加一千个分隔符和一个$符号，但我想使代码更高效。我尝试将它们转换为函数，这样就不必重复我的代码，但是我无法使其正常工作。

df2 = pd.DataFrame({"Category": ["a", "b", "c"], "Num1": [1000, 2000, 3000], "Num2": [4000, 5000, 6000]})

# This works, but I don't want to use those two repetitive lines of code
df2["Num1_2"] = df2.apply(lambda x: "$"+ "{:,}".format(x['Num1']), axis=1)
df2["Num2_2"] = df2.apply(lambda x: "$"+ "{:,}".format(x['Num2']), axis=1)

# I tried writing this function instead
def convert(column):
    "$"+ "{:,}".format(column)
    return column

# This returns an error: KeyError: "None of [Index(['Num1', 'Num2'], dtype='object')] are in the [columns]"
df[["Num1_2", "Num2_2"]] = df[["Num1", "Num2"]].apply(convert, axis = "columns")

Answer 1

在这种情况下，您要使用.applymap()，因为您要更改每个单独的项目：

def convert(v):
    return "$"+ "{:,}".format(v)

df2[["Num1_3", "Num2_3"]] = df2[["Num1", "Num2"]].applymap(convert)

print(df2)

收益：

  Category  Num1  Num2  Num1_2  Num2_2  Num1_3  Num2_3
0        a  1000  4000  $1,000  $4,000  $1,000  $4,000
1        b  2000  5000  $2,000  $5,000  $2,000  $5,000
2        c  3000  6000  $3,000  $6,000  $3,000  $6,000

熊猫-将Lambda转化为功能

1 个答案: