僅供參考，性能/速度對(duì)于這個(gè)問題并不重要。我有一個(gè)名為cost_table...的現(xiàn)有熊貓數(shù)據(jù)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cost_source根據(jù)多個(gè)字段中的值添加一個(gè)字段。google 上出現(xiàn)的大多數(shù)答案都涉及列表推導(dǎo)式或三元運(yùn)算符，但那些僅包含基于一列中的值的邏輯。例如，cost_table['cost_source'] = ['map' if qty > 0 else None for qty in cost_table['qty']]這基于一列中的值起作用，但我不知道如何擴(kuò)展它以包含多列中的邏輯（或者甚至可能？）。它似乎也不是一個(gè)非?？勺x/可維護(hù)的解決方案。我嘗試使用for in帶有if elif語句的循環(huán)，但 in 的值cost_table['cost_source']保持不變并且適用None于所有行。但是，如果我在循環(huán)中打印每一行，則row['cost_source']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'] = Nonefor 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 expectedprint(cost_table)哪個(gè)輸出...
2 回答
catspeake
TA貢獻(xiàn)1111條經(jīng)驗(yàn) 獲得超0個(gè)贊
使用df.apply(lambda x: fun(x), 但帶有參數(shù)axis=1，因此 lambda 函數(shù)逐行應(yīng)用（默認(rèn)為逐列）。
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)
（我還糾正了一個(gè)不一致的地方：在數(shù)據(jù)procents中大概應(yīng)該乘以 100）
30秒到達(dá)戰(zhàn)場(chǎng)
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超6個(gè)贊
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
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