我正在嘗試在下面的“父”列和“子”列列表中的每個(gè)元素上運(yùn)行幾段代碼。我的目標(biāo)結(jié)果在“ resultsX ”列中：第一個(gè)代碼（顯示在 results1 列中）片段只需要兩個(gè)字符串并獲取每個(gè)字符串中的唯一字符總數(shù)：>>> len(set('max' + 'andi'))>>> 6第二段代碼（顯示在 results2 列中）計(jì)算兩個(gè)字符串中相同位置的字符數(shù)：>>> sum(1 if c1 == c2 else 0 for c1, c2 in zip('maxwell', ' axwell'))>>> 6我想在“parent”列和“children”列中的每個(gè)元素上運(yùn)行這兩段代碼，分別獲取“ results1 和 results2”列中的內(nèi)容：parent children results1 results20 MAX ['MAX', 'amx', 'akd'] [('MAX',3),('amx',3),('akd',5)] [('MAX',3),('amx',0),('akd',0)]1 Sam ['Sam','sammy','samsam'] [('Sam',3),('sammy',4), ('samsam',3)] [('Sam',3),('sammy',3), ('samsam',3)]2 Larry ['lar','lair','larrylamo'] [('lar',4),('lair',5), ('larrylamo',6)] [('lar',3),('lair',3), ('larrylamo',5)]我在下面嘗試過這段代碼，但運(yùn)行時(shí)出現(xiàn)錯(cuò)誤：def total_chars(reference, value_list): chars = [] for val in value_list: totalChars = len(set(reference + val)) chars.append(val, reference) return totalCharsdf.agg(lambda x: total_chars(df.parent,df.children), axis=1)>>> ValueError: operands could not be broadcast together with shapes (10000,) (4,) 有任何想法嗎？
2 回答
慕容森
TA貢獻(xiàn)1853條經(jīng)驗(yàn) 獲得超18個(gè)贊
我相信你需要
def cust_func(row):
results1 = []
results2 = []
p = row['parent'].lower()
for i in row['children']:
t = i.lower()
results1.append((i, len(set(t + p))))
results2.append((i, sum(int(k==v) for k,v in zip(t, p))))
return pd.Series([results1, results2])
df = pd.DataFrame({"parent": ['Max', 'Sam', 'Larry'], "children": [['MAX', 'amx', 'akd'], ['Sam','sammy','samsam'], ['lar','lair','larrylamo']]})
df[['results1', 'results2']] = df.apply(cust_func, axis=1)
print(df[['results1', 'results2']])
輸出：
results1 results2
0 [(MAX, 3), (amx, 3), (akd, 5)] [(MAX, 3), (amx, 1), (akd, 0)]
1 [(Sam, 3), (sammy, 4), (samsam, 3)] [(Sam, 3), (sammy, 3), (samsam, 3)]
2 [(lar, 4), (lair, 5), (larrylamo, 6)] [(lar, 3), (lair, 3), (larrylamo, 5)]
湖上湖
TA貢獻(xiàn)2003條經(jīng)驗(yàn) 獲得超2個(gè)贊
沒有循環(huán)數(shù)據(jù)幀。只是一個(gè)用于位置明智比較的功能
df = pd.DataFrame([{'parent': 'MAX', 'children': ['MAX', 'amx', 'akd']},
{'parent': 'Sam', 'children': ['Sam', 'sammy', 'samsam']},
{'parent': 'Larry', 'children': ['lar', 'lair', 'larrylamo']}])
def placewise(a1, a2):
s = 0
for i in range(min(len(a1), len(a2))):
if a1[i] == a2[i]: s += 1
return s
df = (df.explode("children")
.assign(
results1=lambda x: x.apply(lambda r: (r["children"],
len(np.unique(np.concatenate([list(r["parent"].upper()),
list(r["children"].upper())])))
), axis=1),
results2=lambda x: x.apply(lambda r: (r["children"], placewise(r["parent"].upper(),
r["children"].upper()
)
), axis=1),
)
.groupby("parent", as_index=False).agg({"children":lambda x: list(x),
"results1":lambda x: list(x),
"results2":lambda x: list(x)})
)
print(df.to_string(index=False))
輸出
parent children results1 results2
Larry [lar, lair, larrylamo] [(lar, 4), (lair, 5), (larrylamo, 6)] [(lar, 3), (lair, 3), (larrylamo, 5)]
MAX [MAX, amx, akd] [(MAX, 3), (amx, 3), (akd, 5)] [(MAX, 3), (amx, 1), (akd, 0)]
Sam [Sam, sammy, samsam] [(Sam, 3), (sammy, 4), (samsam, 3)] [(Sam, 3), (sammy, 3), (samsam, 3)]
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