Question

我有一个看起来像这样的数据框。它有更多的时间轴，直到Time[s].30。

Time[s]    v1   Time[s].1   v2
160.84621   0   160.84808   7
161.14613   0   161.14802   7
161.538245  27  161.540085  7
162.01598   27  162.017865  7
162.31589   27  162.317775  7
162.615855  27  162.617735  7
162.915765  27  162.91765   7
163.21574   27  163.217625  7
163.51569   27  163.517575  7
163.81563   27  163.81751   7
164.11554   27  164.117425  7
164.4155    27  164.41738   9
164.71543   27  164.717315  9
165.015405  27  165.017285  9
165.31532   27  165.317205  9
165.65083   26  165.65272   9
165.95025   26  165.95214   9

我想要一个时间轴Time[s].general，它是所有时间列的合并形式，带有排序值。我已将所有这些列编入索引。

df.set_index(keys=list(file_read.filter(like='Time[s]').columns))

更新

预期产出：

Time[s]      v1     v2
160.84621   0      null 
160.84808   null     7
160.14613   0      null
161.14802   null     7
161.538245  27     null
161.540085  null     7
162.01598   27     null
162.017865  null     7
162.31589   27     null
162.317775  null     7

等等。

更新2：

Time[s]    v1   Time[s].1   v2      Time[s].2   v3
160.84621   0   160.84808   7   158.538395  Active
161.14613   0   161.14802   7   158.538515  Active
161.538245  27  161.540085  7   159.49455   Active
162.01598   27  162.017865  7   162.352395  Locked
162.31589   27  162.317775  7   163.35075   Locked
162.615855  27  162.617735  7   164.350675  Locked
162.915765  27  162.91765   7   165.350655  Locked
163.21574   27  163.217625  7   166.509695  Locked
163.51569   27  163.517575  7   166.509815  Locked
163.81563   27  163.81751   7   167.50086   Locked
164.11554   27  164.117425  7   168.50085   Locked
164.4155    27  164.41738   9   169.500865  Locked
164.71543   27  164.717315  9   171.502655  Standby
165.015405  27  165.017285  9   185.89923   Forward
165.31532   27  165.317205  9   3273.448065 Forward
165.65083   26  165.65272   9   3274.43487  Forward
165.95025   26  165.95214   9   3275.4348   Forward

Answer 1

我认为需要：

b  = df.filter(like='v').columns

d = {x: 'v.{}'.format(i) for i, x in enumerate(b)}
d['Time[s]'] = 'Time[s].0'
print (d)
{'v1': 'v0', 'v2': 'v1', 'Time[s]': 'Time[s].0'}

df = df.rename(columns=d)
L = [x.set_index(x.columns[0]) for i, x in df.groupby(lambda x: x.split('.')[-1], axis=1)]
df = pd.concat(L, axis=1)
print (df.head(10))
             v.0  v.1
160.846210   0.0  NaN
160.848080   NaN  7.0
161.146130   0.0  NaN
161.148020   NaN  7.0
161.538245  27.0  NaN
161.540085   NaN  7.0
162.015980  27.0  NaN
162.017865   NaN  7.0
162.315890  27.0  NaN
162.317775   NaN  7.0

<强> Expanation ：

所有v列的所有rename列用于字对，用于时间戳的值对列。
dict timestamp，也是groupby列
.按列表理解中mean后的columna值，filter和set_index创建索引

编辑：

如果数字值和重复的时间戳聚合是first，如果没有，则按b = df.filter(like='v').columns d = {x: 'v.{}'.format(i) for i, x in enumerate(b)} d['Time[s]'] = 'Time[s].0' print (d) {'v1': 'v0', 'v2': 'v1', 'Time[s]': 'Time[s].0'} df = df.rename(columns=d) L = [x.groupby(x.columns[0]).mean() if np.issubdtype(df[x.columns[1]].dtype, np.number) else x.groupby(x.columns[0]).first() for i, x in df.groupby(df.columns.str.split('.').str[-1], axis=1)] df = pd.concat(L, axis=1) print (df.head(10)) v.0 v.1 v.2 158.538395 NaN NaN Active 158.538515 NaN NaN Active 159.494550 NaN NaN Active 160.846210 0.0 NaN NaN 160.848080 NaN 7.0 NaN 161.146130 0.0 NaN NaN 161.148020 NaN 7.0 NaN 161.538245 27.0 NaN NaN 161.540085 NaN 7.0 NaN 162.015980 27.0 NaN NaN进行聚合：

UNION ALL

在单个数据帧中合并和排序单个列中的所有索引列

1 个答案: