我有 2 個數據框（足夠大）。他們有列：ch_name、時間、值?，F在，df（第一個數據幀）有 160 個通道，但 df1（第二個數據幀）有 161 個通道。如何識別過度的頻道名稱（df 沒有此頻道但 df1 有）。有時通道名稱在 2 個數據幀之間不匹配。最后，我想要 2 個具有相同通道名稱和編號的數據幀（單獨）。我想為兩個數據框制作 160 個頻道。我不想加入或連接它們。 ch_name time value interval0 ADCS.Attitude_-_Omega_Y_deg_s 2019-01-05 00:00:00 -0.059533 60.01 ADCS.Attitude_-_Omega_Y_deg_s 2019-01-05 00:01:00 -0.063133 60.02 ADCS.Attitude_-_Omega_Y_deg_s 2019-01-05 00:02:00 -0.057533 60.03 ADCS.Attitude_-_Omega_Y_deg_s 2019-01-05 00:03:00 -0.053867 60.04 ADCS.Attitude_-_Omega_Y_deg_s 2019-01-05 00:04:00 -0.067067 60.0... ... ... ... ...1435 ADCS.RW3_MsmtValid 2019-01-05 23:55:00 1.000000 60.01436 ADCS.RW3_MsmtValid 2019-01-05 23:56:00 1.000000 60.01437 ADCS.RW3_MsmtValid 2019-01-05 23:57:00 1.000000 60.01438 ADCS.RW3_MsmtValid 2019-01-05 23:58:00 1.000000 60.01439 ADCS.RW3_MsmtValid 2019-01-05 23:59:00 1.000000 60.0
3 回答
HUWWW
TA貢獻1874條經驗 獲得超12個贊
Test data:
df1=pd.DataFrame({"ch_name":["ch1","ch1","ch2","ch3"],"time":pd.date_range("2019-10-12",freq="min",periods=4),"value":[1,2,3,4],"interval":[60,60,60,60]})
df2=pd.DataFrame({"ch_name":["ch3","ch1","ch5","ch3"],"time":pd.date_range("2019-10-11",freq="min",periods=4),"value":[10,20,30,40],"interval":[70,70,70,70]})
df1
ch_name time value interval
0 ch1 2019-10-12 00:00:00 1 60
1 ch1 2019-10-12 00:01:00 2 60
2 ch2 2019-10-12 00:02:00 3 60
3 ch3 2019-10-12 00:03:00 4 60
df2
ch_name time value interval
0 ch3 2019-10-11 00:00:00 10 70
1 ch1 2019-10-11 00:01:00 20 70
2 ch5 2019-10-11 00:02:00 30 70
3 ch3 2019-10-11 00:03:00 40 70
Create a set/array for the common column names:
#ss=set(df1.ch_name) & set(df2.ch_name)
ss= np.intersect1d(df1.ch_name,df2.ch_name)
Create the new data frames:
new_df1=df1[df1.ch_name.isin(ss)]
ch_name time value interval
0 ch1 2019-10-12 00:00:00 1 60
1 ch1 2019-10-12 00:01:00 2 60
3 ch3 2019-10-12 00:03:00 4 60
new_df3=df2[df2.ch_name.isin(ss)]
ch_name time value interval
0 ch3 2019-10-11 00:00:00 10 70
1 ch1 2019-10-11 00:01:00 20 70
3 ch3 2019-10-11 00:03:00 40 70
慕的地10843
TA貢獻1785條經驗 獲得超8個贊
您可以使用設置邏輯執行此操作：
df = pd.DataFrame(columns = ['a1', 'ba1', 'b1', 'c1'])
df1 = pd.DataFrame(columns = ['a1', 'gg', 'b1', 'c2'])
df
a1 ba1 b1 c1
df1
a1 gg b1 c2
要從 df 中刪除 df1 中的列：
df.drop(columns = set(df.columns) - set(df1.columns), inplace=True)
df
a1 b1
蕪湖不蕪
TA貢獻1796條經驗 獲得超7個贊
你試過合并嗎？
我創建了一些虛假數據來向您展示它是如何工作的。
data1 = [['ch1','12:55',512],['ch2','12:57',516], ['ch3','12:57',420],
['ch4','11:48',433],['ch5','10:42',467]]
data2 = [['ch1','12:55',511],['ch2','12:54',519],['ch3','12:51',720],
['ch5','10:45',497]]
import pandas as pd
df1 = pd.DataFrame(data1,columns=('ch_name','time','value'))
df2 = pd.DataFrame(data2,columns=('ch_name','time','value'))
dfResult = pd.merge(df1, df2, on='ch_name', how='inner')
print(dfResult)
ch_name time_x value_x time_y value_y
0 ch1 12:55 512 12:55 511
1 ch2 12:57 516 12:54 519
2 ch3 12:57 420 12:51 720
3 ch5 10:42 467 10:45 497
這里的關鍵是參數 how='inner' 和 on='ch_name'。how='inner' 是你說你想要出現在兩個數據框中的列的方式，而 on = column_name 是你定義要匹配的列的方式。
我希望它有所幫助。
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