我有多个DataFrame，我想做同样的事情。
首先，我创建一个DataFrame列表。所有这些列都有相同的列名为＆＃39;结果＆＃39;。
df_list = [df1,df2,df3]
我想只保留所有DataFrame中的行，并且值为＆＃39;传递＆＃39;所以我在列表中使用for循环：
for df in df_list:
df =df[df['result'] == 'passed']
...这不起作用，不会从每个DataFrame中过滤掉这些值。
如果我单独过滤每一个，那么它确实有效。
df1 =df1[df1['result'] == 'passed']
df2 =df2[df2['result'] == 'passed']
df3 =df3[df3['result'] == 'passed']
答案 0 :(得分：5)
这是因为每次执行像这个df[<whatever>]这样的子集时，您都会返回一个新的数据帧，并将其分配给df循环变量，每次进行下一次迭代时，该变量都会被删除（虽然你确实保留了最后一个）。这类似于切片列表：
>>> list1 = [1,2,3,4]
>>> list2 = [11,12,13,14]
>>> for lyst in list1,list2:
... lyst = lyst[1:-1]
...
>>> list1, list2
([1, 2, 3, 4], [11, 12, 13, 14])
>>> lyst
[12, 13]
通常，如果要实际修改列表，则需要使用mutator方法。同样，对于数据框，您可以在索引器上使用赋值，例如.loc/.ix/.iloc/等与.dropna方法结合使用，小心传递inplace=True参数。假设我有三个数据帧，我想只保留第二列为正的行：
In [11]: df1
Out[11]:
0 1 2 3
0 0.957288 -0.170286 0.406841 -3.058443
1 1.762343 -1.837631 -0.867520 1.666193
2 0.618665 0.660312 -1.319740 -0.024854
3 -2.008017 -0.445997 -0.028739 -0.227665
4 0.638419 -0.271300 -0.918894 1.524009
5 0.957006 1.181246 0.513298 0.370174
6 0.613378 -0.852546 -1.778761 -1.386848
7 -1.891993 -0.304533 -1.427700 0.099904
In [12]: df2
Out[12]:
0 1 2 3
0 -0.521018 0.407258 -1.167445 -0.363503
1 -0.879489 0.008560 0.224466 -0.165863
2 0.550845 -0.102224 -0.575909 -0.404770
3 -1.171828 -0.912451 -1.197273 0.719489
4 -0.887862 1.073306 0.351835 0.313953
5 -0.517824 -0.096929 -0.300282 0.716020
6 -1.121527 0.183219 0.938509 0.842882
7 0.003498 -2.241854 -1.146984 -0.751192
In [13]: df3
Out[13]:
0 1 2 3
0 0.240411 0.795132 -0.305770 -0.332253
1 -1.162097 0.055346 0.094363 -1.254859
2 -0.493466 -0.717872 1.090417 -0.591872
3 1.021246 -0.060453 -0.013952 0.304933
4 -0.859882 -0.947950 0.562609 1.313632
5 0.917199 1.186865 0.354839 -1.771787
6 -0.694799 -0.695505 -1.077890 -0.880563
7 1.088068 -0.893466 -0.188419 -0.451623
In [14]: for df in df1, df2, df3:
....: df.loc[:,:] = df.loc[df[1] > 0,:]
....: df.dropna(inplace = True,axis =0)
....:
In [15]: df1
dfOut[15]:
0 1 2 3
2 0.618665 0.660312 -1.319740 -0.024854
5 0.957006 1.181246 0.513298 0.370174
In [16]: df2
Out[16]:
0 1 2 3
0 -0.521018 0.407258 -1.167445 -0.363503
1 -0.879489 0.008560 0.224466 -0.165863
4 -0.887862 1.073306 0.351835 0.313953
6 -1.121527 0.183219 0.938509 0.842882
In [17]: df3
Out[17]:
0 1 2 3
0 0.240411 0.795132 -0.305770 -0.332253
1 -1.162097 0.055346 0.094363 -1.254859
5 0.917199 1.186865 0.354839 -1.771787
我认为我找到了一种更好的方法，只需使用.drop方法。
In [21]: df1
Out[21]:
0 1 2 3
0 -0.804913 -0.481498 0.076843 1.136567
1 -0.457197 -0.903681 -0.474828 1.289443
2 -0.820710 1.610072 0.175455 0.712052
3 0.715610 -0.178728 -0.664992 1.261465
4 -0.297114 -0.591935 0.487698 0.760450
5 1.035231 -0.108825 -1.058996 0.056320
6 1.579931 0.958331 -0.653261 -0.171245
7 0.685427 1.447411 0.001002 0.241999
In [22]: df2
Out[22]:
0 1 2 3
0 1.660864 0.110002 0.366881 1.765541
1 -0.627716 1.341457 -0.552313 0.578854
2 0.277738 0.128419 -0.279720 -1.197483
3 -1.294724 1.396698 0.108767 1.353454
4 -0.379995 0.215192 1.446584 0.530020
5 0.557042 0.339192 -0.105808 -0.693267
6 1.293941 0.203973 -3.051011 1.638143
7 -0.909982 1.998656 -0.057350 2.279443
In [23]: df3
Out[23]:
0 1 2 3
0 -0.002327 -2.054557 -1.752107 -0.911178
1 -0.998328 -1.119856 1.468124 -0.961131
2 -0.048568 0.373192 -0.666330 0.867719
3 0.533597 -1.222963 0.119789 -0.037949
4 1.203075 -0.773511 0.475809 1.352943
5 -0.984069 -0.352267 -0.313516 0.138259
6 0.114596 0.354404 2.119963 -0.452462
7 -1.033029 -0.787237 0.479321 -0.818260
In [25]: for df in df1,df2,df3:
....: df.drop(df.index[df[1] < 0],axis=0,inplace=True)
....:
In [26]: df1
Out[26]:
0 1 2 3
2 -0.820710 1.610072 0.175455 0.712052
6 1.579931 0.958331 -0.653261 -0.171245
7 0.685427 1.447411 0.001002 0.241999
In [27]: df2
Out[27]:
0 1 2 3
0 1.660864 0.110002 0.366881 1.765541
1 -0.627716 1.341457 -0.552313 0.578854
2 0.277738 0.128419 -0.279720 -1.197483
3 -1.294724 1.396698 0.108767 1.353454
4 -0.379995 0.215192 1.446584 0.530020
5 0.557042 0.339192 -0.105808 -0.693267
6 1.293941 0.203973 -3.051011 1.638143
7 -0.909982 1.998656 -0.057350 2.279443
In [28]: df3
Out[28]:
0 1 2 3
2 -0.048568 0.373192 -0.666330 0.867719
6 0.114596 0.354404 2.119963 -0.452462
当然更快：
In [8]: timeit.Timer(stmt="df.loc[:,:] = df.loc[df[1] > 0, :];df.dropna(inplace = True,axis =0)", setup="import pandas as pd,numpy as np; df = pd.DataFrame(np.random.random((8,4)))").timeit(10000)
Out[8]: 23.69621358400036
In [9]: timeit.Timer(stmt="df.drop(df.index[df[1] < 0],axis=0,inplace=True)", setup="import pandas as pd,numpy as np; df = pd.DataFrame(np.random.random((8,4)))").timeit(10000)
Out[9]: 11.476448250003159
答案 1 :(得分：0)
如果您的多列包含通过的条件，如超过 40 则通过否则失败
col_list_contains_passed= ['result1','result2']
df_list = [df1,df2,df3]
for d in df_list:
for c in col_list:
d[c]=np.where(d[c]>=40,'Passed','failed')#you can put your condition
然后您可以按条件过滤单个数据框
df1=df1[df1['col_name']=='xyz]
如果 for 循环有效，我会更新答案。待续...