Question

仅供参考，对于这个问题，性能/速度不重要。

我有一个名为cost_table的现有熊猫数据框...

+----------+---------+------+-------------------------+-----------------+
| material | percent | qty  | price_control_indicator | acct_assign_cat |
+----------+---------+------+-------------------------+-----------------+
| abc111   | 1.00    |   50 | v                       | #               |
| abc222   | 0.25    | 2000 | s                       | #               |
| xyz789   | 0.45    |    0 | v                       | m               |
| def456   | 0.9     |    0 | v                       | #               |
| 123xyz   | 0.2     |    0 | v                       | m               |
| lmo888   | 0.6     |    0 | v                       | m               |
+----------+---------+------+-------------------------+-----------------+

我需要基于多个字段中的值添加字段cost_source。

大多数出现在Google上的答案都涉及列表理解或三元运算符，但这些答案仅包括基于一列中值的逻辑。例如，

cost_table['cost_source'] = ['map' if qty > 0 else None for qty in cost_table['qty']]

这是基于一列中的值工作的，但是我不知道如何扩展它以在多列中包含逻辑（或者是否有可能？）。它似乎也不是一个易于阅读/可维护的解决方案。

我尝试将for in循环与if elif语句一起使用，但是cost_table['cost_source']中的值保持不变，并且对于所有行都是None。但是，如果我在循环中打印每一行，则row['cost_source']会具有所需的值。

d = {
  'material': ['abc111', 'abc222', 'xyz789', 'def456', '123xyz', 'lmo888'],
  'percent': [1, .25, .45, .9, .2, .6],
  'qty': [50, 2000, 0, 0, 0, 0],
  'price_control_indicator': ['v', 's','v', 'v', 'v', 'v'],
  'acct_assign_cat': ['#', '#', 'm', '#', 'm', 'm']
}

cost_table = pd.DataFrame(data=d)

cost_table['cost_source'] = None

for index, row in cost_table.iterrows():
  if (row['qty'] > 0) or (row['price_control_indicator'] == "s") or (row['acct_assign_cat'] == "#"):
    row['cost_source'] = "map"
  elif (row['percent'] >= 40) and (row['acct_assign_cat'] == "m"):
    row['cost_source'] = "vendor"
  else:
    row['cost_source'] = None

  print(row['cost_source']) # outputs map, vendor, or None as expected

print(cost_table)

哪个输出...

+----------+---------+------+-------------------------+-----------------+-------------+
| material | percent | qty  | price_control_indicator | acct_assign_cat | cost_source |
+----------+---------+------+-------------------------+-----------------+-------------+
| abc111   | 1.00    |   50 | v                       | #               | None        |
| abc222   | 0.25    | 2000 | s                       | #               | None        |
| xyz789   | 0.45    |    0 | v                       | m               | None        |
| def456   | 0.9     |    0 | v                       | #               | None        |
| 123xyz   | 0.2     |    0 | v                       | m               | None        |
| lmo888   | 0.6     |    0 | v                       | m               | None        |
+----------+---------+------+-------------------------+-----------------+-------------+

这是我想要的结果...

+----------+---------+------+-------------------------+-----------------+-------------+
| material | percent | qty  | price_control_indicator | acct_assign_cat | cost_source |
+----------+---------+------+-------------------------+-----------------+-------------+
| abc111   | 1.00    |   50 | v                       | #               | map         |
| abc222   | 0.25    | 2000 | s                       | #               | map         |
| xyz789   | 0.45    |    0 | v                       | m               | vendor      |
| def456   | 0.9     |    0 | v                       | #               | map         |
| 123xyz   | 0.2     |    0 | v                       | m               | None        |
| lmo888   | 0.6     |    0 | v                       | m               | vendor      |
+----------+---------+------+-------------------------+-----------------+-------------+

Answer 1

如@bazinga所述，使用df.apply(lambda x: fun(x)，但使用参数axis=1，因此lambda函数将逐行应用（默认为逐列）。

d = {
  'material': ['abc111', 'abc222', 'xyz789', 'def456', '123xyz', 'lmo888'],
  'percent': [100, 25, 45, 90, 20, 60],
  'qty': [50, 2000, 0, 0, 0, 0],
  'price_control_indicator': ['v', 's','v', 'v', 'v', 'v'],
  'acct_assign_cat': ['#', '#', 'm', '#', 'm', 'm']
}

cost_table = pd.DataFrame(data=d)

def process_row(row):
    if (row['qty'] > 0) or (row['price_control_indicator'] == "s") or (row['acct_assign_cat'] == "#"):
        return "map"
    elif (row['percent'] >= 40) and (row['acct_assign_cat'] == "m"):
        return "vendor"
    else:
        return None

cost_table['cost_source'] = cost_table.apply(lambda row: process_row(row), axis=1)

print(cost_table)

（我还纠正了一个不一致的问题：在数据procents中应该乘以100）

Answer 2

如果您想使用np.select

cond1 = cost_table.qty.gt(0) | cost_table.price_control_indicator.eq('s') | cost_table.acct_assign_cat.eq('#')
cond2 = cost_table.percent.ge(0.4) & cost_table.acct_assign_cat.eq('m')
cost_table['cost_source'] = np.select([cond1, cond2], ['map', 'vendor'], default='None')
print(cost_table)

  material  percent   qty price_control_indicator acct_assign_cat cost_source
0   abc111     1.00    50                       v               #         map
1   abc222     0.25  2000                       s               #         map
2   xyz789     0.45     0                       v               m      vendor
3   def456     0.90     0                       v               #         map
4   123xyz     0.20     0                       v               m        None
5   lmo888     0.60     0                       v               m      vendor

根据多个列中的值创建新的数据框列

2 个答案: