Question

我正在将一个函数转换为循环集合的pandas，并根据条件和运行总计更新每个值。该函数看起来像这样

def calculate_value():
    cumulative_amount = 0

    for row in rows:
        if row['amount'] < 0:
            return 0

        amount = 0

        if row['kind'] == 'A':
            amount = row['amount'] * row['input_amount']
        elif row['kind'] == 'B':
            amount = row['input_amount'] - cumulative_amount
        elif row['kind'] == 'C':
            amount = row['amount']

        cumulative_amount += amount
        row['result'] = amount

        if row['kind'] == 'B':
            break

    return rows

基本上，遍历所有行，并添加result值。但是此result可能取决于累计运行总计。此外，如果我们达到某个值（row['kind'] == 'B'），我们应该打破并停止处理新行。

将其转换为pandas时，似乎我应该使用apply。到目前为止，我有以下代码，其中几乎有效，但当我尝试使用cumulative_amount获取shift(-1)时，它总是以nan的形式返回。< / p>

在熊猫中做到这一点的最佳方式是什么？

def calculate_value(row: Series):
    if row['amount'] < 0 or row.shift(-1)['kind'] == 'B':
        row['cumulative_amount'] = 0
        row['result'] = 0
        return row

    amount = 0

    if np.isnan(row.shift(-1)['cumulative_amount']):
        cumulative_amount = 0
    else:
        cumulative_amount = row.shift(-1)['cumulative_amount']

    if row['kind'] == 'A':
        amount = row['amount'] * row['input_amount']
    elif row['kind'] == 'B':
        amount = row['input_amount'] - cumulative_amount
    elif row['kind'] == 'C':
        amount = row['amount']

    row['cumulative_amount'] = amount + cumulative_amount
    row['result'] = amount
    return row

df['cumulative_amount'] = 0
new_df = df.apply(lambda x: calculate_value(x), axis=1)

输入和所需输出的示例是

df = pd.DataFrame({
    'kind': {1: 'C', 2: 'E', 3: 'A', 4: 'A', 5: 'B', 6: 'C'},
    'amount': {1: -800, 2: 100, 3: 0.5, 4: 0.5, 5: 0, 6: 200},
    'input_amount': {1: 800, 2: 800, 3: 800, 4: 800, 5: 800, 6: 800}
})

   amount  input_amount kind  cumulative_amount  result
1  -800.0           800    C                0.0     0.0
2   100.0           800    E                0.0     0.0
3     0.5           800    A              400.0   400.0
4     0.5           800    A              800.0   400.0
5     0.0           800    B              800.0     0.0
6   200.0           800    C              800.0     0.0

Answer 1

如果我理解正确，只有result种类'B'取决于其他行。所以你可以先做其他事情开始：

df['result'] = 0.

a = (df.kind == 'A') & (df.amount >= 0) 
c = (df.kind == 'C') & (df.amount >= 0)

df.loc[a, 'result'] = df.loc[a, 'amount'] * df.loc[a, 'input_amount']
df.loc[c, 'result'] = df.loc[c, 'amount']

做cumsum：

df['cumulative_amount'] = df.result.cumsum()

更正'cumulative_amount'的值（对于所有类型'B'的出现次数）：

df.loc[(df.kind == 'B'), 'result'] = df.loc[(df.kind == 'B'), 'input_amount'].values - df.loc[(df.kind.shift(-1) == 'B'), 'cumulative_amount'].values

在第一次出现'result'后更正'cumulative_amount'和'B'的值：

df.loc[(df.kind == 'B').cumsum().shift() > 0, 'result'] = 0
# (df.kind == 'B').cumsum().shift() is a running count of the number of B's encountered prior to the row index, 
# so you want to 'stop' once this number is no longer zero
# You could of course do this more simply by figuring out which position in the index has the first B, 
# then using .ix or .iloc, but it's actually longer to type out.

df['cumulative_amount'] = df.result.cumsum() # Once more, because we've changed the value of results below B.

Pandas适用于第二列

1 个答案: