我有一段串行代码,可以做这样的事情
if( ! variable )
{
do some initialization here
variable = true;
}
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据我所知,这在串行中完美无缺,只执行一次.什么原子操作在CUDA中是正确的?
我有一个带有条件检查的循环,如果条件为真但仍然继续循环,有没有办法可以退出当前迭代.
dummy = ['One','Two','Three','Four','Five']
for i in dummy:
print('Stage 1')
if i == 'Three':
break
print('Stage 2')
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这段代码给了我:
Stage 1
Stage 2
Stage 1
Stage 2
Stage 1
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但我想这样:
Stage 1
Stage 2
Stage 1
Stage 2
--------
Stage 1
--------
Stage 1
Stage 2
Stage 1
Stage 2
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突出显示的行显示它跳过元素"Three"的第二个print语句.
谢谢
我有一个这样的DataFrame
dict_ = {'Date':['2018-01-01','2018-01-02','2018-01-03','2018-01-04','2018-01-05'],'Col1':[1,2,3,4,5],'Col2':[1.1,1.2,1.3,1.4,1.5],'Col3':[0.33,0.98,1.54,0.01,0.99]}
df = pd.DataFrame(dict_, columns=dict_.keys())
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然后,我计算列之间的皮尔逊相关性,并过滤掉相关于我的阈值0.95以上的列
def trimm_correlated(df_in, threshold):
df_corr = df_in.corr(method='pearson', min_periods=1)
df_not_correlated = ~(df_corr.mask(np.eye(len(df_corr), dtype=bool)).abs() > threshold).any()
un_corr_idx = df_not_correlated.loc[df_not_correlated[df_not_correlated.index] == True].index
df_out = df_in[un_corr_idx]
return df_out
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产生
uncorrelated_factors = trimm_correlated(df, 0.95)
print uncorrelated_factors
Col3
0 0.33
1 0.98
2 1.54
3 0.01
4 0.99
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到目前为止,我对结果感到满意,但我想保留每个相关对中的一列,因此在上面的示例中,我想包含Col1或Col2。得到某物 像这样
Col1 Col3
0 1 0.33
1 2 0.98
2 3 1.54
3 4 0.01
4 5 0.99
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另外,我还可以做进一步的评估来确定保留哪些相关列?
谢谢