我有一些看起來像這樣的資料：
import pandas as pd
fruits = ['apple', 'pear', 'peach']
df = pd.DataFrame({'col1':['i want an apple', 'i hate pears', 'please buy a peach and an apple', 'I want squash']})
print(df.head())
col1
0 i want an apple
1 i hate pears
2 please buy a peach and an apple
3 I want squash
我需要一個解決方案，為中的每個專案創建一列fruits并給出 1 或 0 值，指示是否col包含該值。理想情況下，輸出將如下所示：
goal_df = pd.DataFrame({'col1':['i want an apple', 'i hate pears', 'please buy a peach and an apple', 'I want squash'],
'apple': [1, 0, 1, 0],
'pear': [0, 1, 0, 0],
'peach': [0, 0, 1, 0]})
print(goal_df.head())
col1 apple pear peach
0 i want an apple 1 0 0
1 i hate pears 0 1 0
2 please buy a peach and an apple 1 0 1
3 I want squash 0 0 0
我試過這個，但沒有用：
for i in fruits:
if df['col1'].str.contains(i):
df[i] = 1
else:
df[i] = 0
uj5u.com熱心網友回復：
items = ['apple', 'pear', 'peach']
for it in items:
df[it] = df['col1'].str.contains(it, case=False).astype(int)
輸出：
>>> df
col1 apple pear peach
0 i want an apple 1 0 0
1 i hate pears 0 1 0
2 please buy a peach and an apple 1 0 1
3 I want squash 0 0 0
uj5u.com熱心網友回復：
使用str.extractall提取的話，那么pd.crosstab：
pattern = f"({'|'.join(fruits)})"
s = df['col1'].str.extractall(pattern)
df[fruits] = (pd.crosstab(s.index.get_level_values(0), s[0].values)
.re_index(index=df.index, columns=fruits, fill_value=0)
)
輸出：
col1 apple pear peach
0 i want an apple 1 0 0
1 i hate pears 0 1 0
2 please buy a peach and an apple 1 0 1
3 I want squash 0 0 0
uj5u.com熱心網友回復：
嘗試：
- 使用獲取所有匹配的水果
str.extractall - 使用
pd.get_dummies來獲得指標值 join到原始資料幀
matches = pd.get_dummies(df["col1"].str.extractall(f"({'|'.join(fruits)})")[0].droplevel(1, 0))
output = df.join(matches.groupby(level=0).sum()).fillna(0)
>>> output
col1 apple peach pear
0 i want an apple 1.0 0.0 0.0
1 i hate pears 0.0 0.0 1.0
2 please buy a peach and an apple 1.0 1.0 0.0
3 I want squash 0.0 0.0 0.0
uj5u.com熱心網友回復：
您可以在下面的蘋果列中使用，并對其他人做同樣的事情
def has_apple(st):
if "apple" in st.lower():
return 1
return 0
df['apple'] = df['col1'].apply(has_apple)
uj5u.com熱心網友回復：
我想到了另一種完全不同的單行：
df[items] = df['col1'].str.findall('|'.join(items)).str.join('|').str.get_dummies('|')
輸出：
>>> df
col1 apple pear peach
0 i want an apple 1 0 0
1 i hate pears 0 0 1
2 please buy a peach and an apple 1 1 0
3 I want squash 0 0 0
uj5u.com熱心網友回復：
嘗試np.where從numpy庫中使用：
fruit = ['apple', 'pear', 'peach']
for i in fruit:
df[i] = np.where(df.col1.str.contains(i), 1, 0)
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