如何將資料框行更改為熊貓中的下一行

我是一個 noob python 用戶，我的目的是得到名字并轉移到下一行

import pandas as pd
import numpy as np
df = pd.DataFrame({"1": ['Alfred', 'car', 'bike','Alex','car'],
                   "2": [np.nan, 'Ford', 'Giant',np.nan,'Toyota'],
                   "3": [pd.NaT, pd.Timestamp("2018-01-01"),
                            pd.Timestamp("2018-07-01"),np.nan,pd.Timestamp("2021-01-01")]})

      1      2       3
0   Alfred  NaN     NaT
1   car    Ford   2018-01-01
2   bike   Giant  2018-07-01
3   Alex    NaN     NaT
4   car    Toyota 2021-01-01

我的目標結果如下

df = pd.DataFrame({"transportation": ['car', 'bike','car'],
                   "Mark": ['Ford', 'Giant','Toyota'],
                   "BuyDate":[pd.Timestamp("2018-01-01"),
                            pd.Timestamp("2018-07-01"),pd.Timestamp("2021-01-01")],
                   "Name":['Alfred','Alfred','Alex']
                   })

  transportation    Mark    BuyDate Name
0   car             Ford    2018-01-01  Alfred
1   bike            Giant   2018-07-01  Alfred
2   car             Toyota  2021-01-01  Alex

我正在嘗試搜索某種方法，但無法解決此問題，感謝您查看我的帖子并提供幫助

uj5u.com熱心網友回復：

您可以使用輔助列然后向前填充來執行此操作：

# rename columns
df.columns = ["transportation", "Mark", "BuyDate"]
# assumption: the rows where "Mark" is NaN defines the name for the following rows
df["is_name"] = df["Mark"].isna()
# create a new column which is NaN everywhere except for the name rows
df["name"] = np.where(df.is_name, df["transportation"], np.nan)
# do a forward fill to extend the names to all rows
df["name"] = df["name"].fillna(method="ffill")
# filter by non-name rows and drop the temporary is_name column
df = df.loc[~df.is_name].drop("is_name", axis=1)
print(df)

Out:
  transportation    Mark    BuyDate    name
1            car    Ford 2018-01-01  Alfred
2           bike   Giant 2018-07-01  Alfred
4            car  Toyota 2021-01-01    Alex

uj5u.com熱心網友回復：

想法是按Mark列向前填充缺失值Name，然后在同一掩碼中過濾行：

df.columns = ["Transportation", "Mark", "BuyDate"]
m = df["Mark"].notna()
df["Name"] = df["transportation"].mask(m).ffill()
df = df[m].reset_index(drop=True)
print(df)
  Transportation    Mark    BuyDate    Name
0            car    Ford 2018-01-01  Alfred
1           bike   Giant 2018-07-01  Alfred
2            car  Toyota 2021-01-01    Alex

uj5u.com熱心網友回復：

您可以使用此管道：

m = df.iloc[:,1].notna()
(df.assign(Name=df.iloc[:,0].mask(m).ffill()) # add new column
   .loc[m] # keep only the columns with info
   # below: rework df to fit output
   .rename(columns={'1': 'transportation', '2': 'Mark', '3': 'BuyDate'})
   .reset_index(drop=True)
)

輸出：

  transportation    Mark    BuyDate    Name
0            car    Ford 2018-01-01  Alfred
1           bike   Giant 2018-07-01  Alfred
2            car  Toyota 2021-01-01    Alex

uj5u.com熱心網友回復：

你可以這樣做：

>>> import pandas as pd
>>> import numpy as np
>>> df = pd.DataFrame({"1": ['Alfred', 'car', 'bike','Alex','car'],
...                    "2": [np.nan, 'Ford', 'Giant',np.nan,'Toyota'],
...                    "3": [pd.NaT, pd.Timestamp("2018-01-01"),
...                             pd.Timestamp("2018-07-01"),np.nan,pd.Timestamp("2021-01-01")]})
>>>
>>> df
        1       2          3
0  Alfred     NaN        NaT
1     car    Ford 2018-01-01
2    bike   Giant 2018-07-01
3    Alex     NaN        NaT
4     car  Toyota 2021-01-01
>>>
>>> new_df = pd.DataFrame(columns=['Transportation', 'Mark', 'BuyDate', 'Name'])
>>>
>>> j = 0
>>> for i in range(1, df.shape[0]):
...   if df.loc[i][1] is np.nan:
...     running_name = df.loc[i][0]
...     continue
...   new_df.loc[j] = list(df.loc[i])   [running_name]
...   j  = 1
...
>>> new_df
  Transportation    Mark    BuyDate    Name
0            car    Ford 2018-01-01  Alfred
1           bike   Giant 2018-07-01  Alfred
2            car  Toyota 2021-01-01    Alex
>>>

標籤：蟒蛇-3.x 熊猫

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