拆分熊猫数据框中的文本列表

Question

我的 pandas 数据框由文本列表组成，我想将文本拆分为 ,

df['c1']=['this is text one','this is text two','this is text three']

我试过了

new = df["c1"].str.split(",", n = 1, expand = True)

但我是 Nan 在新变量上

预期输出

c1='this is text one'
c1='this is text two'
c1='this is text three'

只要拆分列表中的文本，其他输出就可以了。感谢您的帮助 完整代码

import pandas as pd
data={"C1":[["this is text one","this is text two","this is text three"]]}
df=pd.DataFrame(data)
df.head()

Answer 1

您不需要 pandas 来拆分可以用于循环的数组

这就是你需要的

for i in df['C1']:
    for each in i:
        print(each) #outputs each element in the array

Answer 2

使用np.concatenate() 并调用数据框构造函数（因为您已经有了一个字符串列表）：

df_new=pd.DataFrame(np.concatenate(df1.C1),columns=['C1'])
#or pd.DataFrame(df1.C1.values.tolist()).T

---

                   C1
0    this is text one
1    this is text two
2  this is text three

Answer 3

您的问题有点令人困惑-您说您已经有一个文本列表，那么为什么要拆分它？如果你的意思是你有一个数据框，其中的字符串必须用逗号分隔，你可以这样做。

import pandas as pd

df = pd.DataFrame()

df['c1']=['this is the first text, which has some commas, in it',
          'this is text two, which also has commas']

df['lists'] = df['c1'].apply(lambda txt: txt.split(','))

df.head()

运行df['lists'][0] 然后给出['this is the first text', ' which has some commas', ' in it']