给出的答案是正确的，尽管上面有人说你可以使用df。查询('line_race != 0')，这取决于您的问题要快得多。强烈推荐。

给出的答案是正确的，尽管上面有人说你可以使用df。查询('line_race != 0')，这取决于您的问题要快得多。强烈推荐。

另一种方法。可能不是最有效的方法，因为代码看起来比其他答案中提到的代码更复杂一些，但仍然是做相同事情的另一种方法。

  df = df.drop(df[df['line_race']==0].index)

我编译并运行我的代码。这是精确的代码。你可以自己试试。

data = pd.read_excel('file.xlsx')

如果你在列名中有任何特殊字符或空格，你可以像给定的代码中那样写:

data = data[data['expire/t'].notnull()]
print (date)

如果只有一个字符串列名，没有任何空格或特殊字符 字符你可以直接访问它。

data = data[data.expire ! = 0]
print (date)

有很多方法可以实现这一点。将在下面留下各种选项，可以使用，这取决于一个用例的特殊性。

可以认为OP的数据帧存储在变量df中。

---

选项1

对于OP的情况，考虑到唯一值为0的列是line_race，下面将完成工作

 df_new = df[df != 0].dropna()
 
[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

然而，由于情况并非总是如此，建议检查以下选项，其中将指定列名。

---

选项2

tshauck的方法最终比选项1更好，因为它可以指定列。然而，根据想要引用列的方式，还有其他的变化:

例如，使用数据框架中的位置

df_new = df[df[df.columns[2]] != 0]

或者通过如下显式地指示列

df_new = df[df['line_race'] != 0]

也可以遵循相同的登录，但使用自定义lambda函数，例如

df_new = df[df.apply(lambda x: x['line_race'] != 0, axis=1)]

[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

---

选项3

使用pandas.Series.map和自定义lambda函数

df_new = df['line_race'].map(lambda x: x != 0)

[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

---

选项4

使用pandas. datafframe .drop，如下所示

df_new = df.drop(df[df['line_race'] == 0].index)

[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

---

选择5

使用pandas.DataFrame.query如下所示

df_new = df.query('line_race != 0')

[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

---

选择6

使用pandas.DataFrame.drop和pandas.DataFrame.query如下所示

df_new = df.drop(df.query('line_race == 0').index)

[Out]:
     line_date  daysago  line_race  rating        rw    wrating
0   2007-03-31       62       11.0      56  1.000000  56.000000
1   2007-03-10       83       11.0      67  1.000000  67.000000
2   2007-02-10      111        9.0      66  1.000000  66.000000
3   2007-01-13      139       10.0      83  0.880678  73.096278
4   2006-12-23      160       10.0      88  0.793033  69.786942
5   2006-11-09      204        9.0      52  0.636655  33.106077
6   2006-10-22      222        8.0      66  0.581946  38.408408
7   2006-09-29      245        9.0      70  0.518825  36.317752
8   2006-09-16      258       11.0      68  0.486226  33.063381
9   2006-08-30      275        8.0      72  0.446667  32.160051
10  2006-02-11      475        5.0      65  0.164591  10.698423

---

选择7

如果对输出没有强烈的意见，可以使用numpy.select的向量化方法

df_new = np.select([df != 0], [df], default=np.nan)

[Out]:
[['2007-03-31' 62 11.0 56 1.0 56.0]
 ['2007-03-10' 83 11.0 67 1.0 67.0]
 ['2007-02-10' 111 9.0 66 1.0 66.0]
 ['2007-01-13' 139 10.0 83 0.880678 73.096278]
 ['2006-12-23' 160 10.0 88 0.793033 69.786942]
 ['2006-11-09' 204 9.0 52 0.636655 33.106077]
 ['2006-10-22' 222 8.0 66 0.581946 38.408408]
 ['2006-09-29' 245 9.0 70 0.518825 36.317752]
 ['2006-09-16' 258 11.0 68 0.486226 33.063381]
 ['2006-08-30' 275 8.0 72 0.446667 32.160051]
 ['2006-02-11' 475 5.0 65 0.164591 10.698423]]

这也可以转换为一个数据框架

df_new = pd.DataFrame(df_new, columns=df.columns)

[Out]:
     line_date daysago line_race rating        rw    wrating
0   2007-03-31      62      11.0     56       1.0       56.0
1   2007-03-10      83      11.0     67       1.0       67.0
2   2007-02-10     111       9.0     66       1.0       66.0
3   2007-01-13     139      10.0     83  0.880678  73.096278
4   2006-12-23     160      10.0     88  0.793033  69.786942
5   2006-11-09     204       9.0     52  0.636655  33.106077
6   2006-10-22     222       8.0     66  0.581946  38.408408
7   2006-09-29     245       9.0     70  0.518825  36.317752
8   2006-09-16     258      11.0     68  0.486226  33.063381
9   2006-08-30     275       8.0     72  0.446667  32.160051
10  2006-02-11     475       5.0     65  0.164591  10.698423

---

至于最有效的解决方案，这将取决于人们如何衡量效率。假设要度量执行时间，可以使用time.perf_counter()。

如果测量上述所有选项的执行时间，就会得到以下结果

       method                   time
0    Option 1 0.00000110000837594271
1  Option 2.1 0.00000139995245262980
2  Option 2.2 0.00000369996996596456
3  Option 2.3 0.00000160001218318939
4    Option 3 0.00000110000837594271
5    Option 4 0.00000120000913739204
6    Option 5 0.00000140001066029072
7    Option 6 0.00000159995397552848
8    Option 7 0.00000150001142174006

但是，这可能会根据所使用的数据框架、需求(比如硬件)等而改变。

---

注:

There are various suggestions on using inplace=True. Would suggest reading this: https://stackoverflow.com/a/59242208/7109869 There are also some people with strong opinions on .apply(). Would suggest reading this: When should I (not) want to use pandas apply() in my code? If one has missing values, one might want to consider as well pandas.DataFrame.dropna. Using the option 2, it would be something like df = df[df['line_race'] != 0].dropna() There are additional ways to measure the time of execution, so I would recommend this thread: How do I get time of a Python program's execution?

虽然前面的答案与我将要做的几乎类似，但使用索引方法并不需要使用另一个索引方法.loc()。它可以以类似但精确的方式完成

df.drop(df.index[df['line_race'] == 0], inplace = True)