问题描述

我有一个 pd.DataFrame，看起来像:

I have a a pd.DataFrame that looks like:

我想对值创建一个截止值以将它们推入二进制数字，在这种情况下我的截止值是 0.85.我希望生成的数据框看起来像:

I want to create a cutoff on the values to push them into binary digits, my cutoff in this case is 0.85. I want the resulting dataframe to look like:

我为此编写的脚本很容易理解，但对于大型数据集，它效率低下.我确信 Pandas 有一些方法可以处理这些类型的转换.

The script I wrote to do this is easy to understand but for large datasets it is inefficient. I'm sure Pandas has some way of taking care of these types of transformations.

有人知道使用阈值将浮点数列转换为整数列的有效方法吗?

我做这种事情的方式极其天真:

My extremely naive way of doing such a thing:

DF_test = pd.DataFrame(np.array([list("abcde"),list("pqrst"),[0.12,0.23,0.93,0.86,0.33]]).T,columns=["c1","c2","value"])
DF_want = pd.DataFrame(np.array([list("abcde"),list("pqrst"),[0,0,1,1,0]]).T,columns=["c1","c2","value"])


threshold = 0.85

#Empty dataframe to append rows
DF_naive = pd.DataFrame()
for i in range(DF_test.shape[0]):
    #Get first 2 columns
    first2cols = list(DF_test.ix[i][:-1])
    #Check if value is greater than threshold
    binary_value = [int((bool(float(DF_test.ix[i][-1]) > threshold)))]
    #Create series object
    SR_row = pd.Series( first2cols + binary_value,name=i)
    #Add to empty dataframe container
    DF_naive = DF_naive.append(SR_row)
#Relabel columns
DF_naive.columns = DF_test.columns
DF_naive.head()
#the sample DF_want

推荐答案

您可以使用 np.where 根据布尔条件设置所需的值:

You can use np.where to set your desired value based on a boolean condition:

In [18]:
DF_test['value'] = np.where(DF_test['value'] > threshold, 1,0)
DF_test

Out[18]:
  c1 c2  value
0  a  p      0
1  b  q      0
2  c  r      1
3  d  s      1
4  e  t      0

请注意，由于您的数据是一个异构 np 数组，因此值"列包含字符串而不是浮点数:

Note that because your data is a heterogenous np array the 'value' column contains strings rather than floats:

In [58]:
DF_test.iloc[0]['value']

Out[58]:
'0.12'

因此，您需要先将 dtype 转换为 float:DF_test['value'] = DF_test['value'].astype(float)

So you'll need to convert the dtype to float first: DF_test['value'] = DF_test['value'].astype(float)

您可以比较时间:

In [16]:
%timeit np.where(DF_test['value'] > threshold, 1,0)
1000 loops, best of 3: 297 s per loop

In [17]:
%%timeit
DF_naive = pd.DataFrame()
for i in range(DF_test.shape[0]):
    #Get first 2 columns
    first2cols = list(DF_test.ix[i][:-1])
    #Check if value is greater than threshold
    binary_value = [int((bool(float(DF_test.ix[i][-1]) > threshold)))]
    #Create series object
    SR_row = pd.Series( first2cols + binary_value,name=i)
    #Add to empty dataframe container
    DF_naive = DF_naive.append(SR_row)
10 loops, best of 3: 39.3 ms per loop

np.where 版本快了 100 倍以上，诚然您的代码做了很多不必要的事情，但您明白了

the np.where version is over 100x faster, admittedly your code is doing a lot of unnecessary stuff but you get the point

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