如何将具有重复列名的行切片并在ord中堆叠该行

2024-04-23 18:49:10 发布

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我有一个如图所示的数据帧,我想把它转换成多行而不改变顺序。你知道吗

  RESP HR SPO2 PULSE
1  46  122  0    0   
2  46  122  0    0   
3
4

Tags: 数据顺序hrresppulsespo2
2条回答

一种可能的解决方案是使用^{},列长度的唯一必要模是0(因此可以将所有数据转换为4列DataFrame):

df1 = pd.Dataframe(df.values.reshape(-1, 4), columns=['RESP','HR','SPO2','PULSE'])
df1['RESP1'] = df['RESP'].shift(-1)

通用数据解决方案:

a = '46 122 0 0 46 122 0 0 45 122 0 0 45 122 0'.split()
df = pd.DataFrame([a]).astype(int)
print (df)
    0    1  2  3   4    5  6  7   8    9  10  11  12   13  14
0  46  122  0  0  46  122  0  0  45  122   0   0  45  122   0

#flatten values
a = df.values.ravel()
#number of new columns
N = 4
#array filled by NaNs for possible add NaNs to end of last row
arr = np.full(((len(a) - 1)//N + 1)*N, np.nan)
#fill array by flatten values
arr[:len(a)] = a
#reshape to new DataFrame (last value is NaN)
df1 = pd.DataFrame(arr.reshape((-1, N)), columns=['RESP','HR','SPO2','PULSE'])
#new column with shifting first col
df1['RESP1'] = df1['RESP'].shift(-1)
print(df1)
   RESP     HR  SPO2  PULSE  RESP1
0  46.0  122.0   0.0    0.0   46.0
1  46.0  122.0   0.0    0.0   45.0
2  45.0  122.0   0.0    0.0   45.0
3  45.0  122.0   0.0    NaN    NaN

另一种方法是groupby

df = pd.DataFrame(np.random.arange(12), columns=list('abcd'*3))

new_df = pd.concat((x.stack().reset_index(drop=True)
                     .rename(k) for k,x in df.groupby(df.columns, axis=1)), 
                    axis=1)

new_df = (new_df.assign(a1=lambda x: x['a'].shift(-1))
                .rename(columns={'a1':'a'})
         )

输出:

   a  b   c   d    a
0  0  1   2   3  4.0
1  4  5   6   7  8.0
2  8  9  10  11  NaN

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