我有一个熊猫数据框列，看起来像：
Out[67]:
0 ["cheese", "milk...
1 ["yogurt", "cheese...
2 ["cheese", "cream"...
3 ["milk", "cheese"...
现在，最终我希望将其作为一个扁平列表，但是在尝试扁平化时，我注意到熊猫将
["cheese", "milk", "cream"]视为str而不是list我如何将其展平，所以最终得到：
["cheese", "milk", "yogurt", "cheese", "cheese"...]
[编辑]
因此，下面给出的答案似乎是：
s = pd.Series(["['cheese', 'milk']", "['yogurt', 'cheese']", "['cheese', 'cream']"])s = s.str.strip("[]")
df = s.str.split(',', expand=True)
df = df.applymap(lambda x: x.replace("'", '').strip())
l = df.values.flatten()
print (l.tolist())
很好，已回答问题，已接受回答，但这对我来说是一个相当微妙的解决方案。
最佳答案
您可以使用numpy.flatten，然后使用平面嵌套的lists-see：
print df
a
0 [cheese, milk]
1 [yogurt, cheese]
2 [cheese, cream]
print df.a.values
[[['cheese', 'milk']]
[['yogurt', 'cheese']]
[['cheese', 'cream']]]
l = df.a.values.flatten()
print l
[['cheese', 'milk'] ['yogurt', 'cheese'] ['cheese', 'cream']]
print [item for sublist in l for item in sublist]
['cheese', 'milk', 'yogurt', 'cheese', 'cheese', 'cream']
编辑：
你可以试试：
import pandas as pd
s = pd.Series(["['cheese', 'milk']", "['yogurt', 'cheese']", "['cheese', 'cream']"])
#remove []
s = s.str.strip('[]')
print s
0 'cheese', 'milk'
1 'yogurt', 'cheese'
2 'cheese', 'cream'
dtype: object
df = s.str.split(',', expand=True)
#remove ' and strip empty string
df = df.applymap(lambda x: x.replace("'", '').strip())
print df
0 1
0 cheese milk
1 yogurt cheese
2 cheese cream
l = df.values.flatten()
print l.tolist()
['cheese', 'milk', 'yogurt', 'cheese', 'cheese', 'cream']