评估重复样本在其字段中是否有不同的数据，以及是否复制数据？

Question

我想评估一个样本及其复制品（以2结尾）是否在他们的年龄、家族史和诊断字段中输入了数据。如果一个示例有条目，而它的副本没有（所有的“-”条目），那么我想将条目从示例复制到duplicates字段。评估应该以另一种方式进行：如果副本有条目，而示例没有条目，则将它们复制到示例字段。你知道吗

基本上，我希望输入的_-df看起来像所需的_-df（如下所示）。你知道吗

input_df = pd.DataFrame(columns=['Sample', 'Date','Age', 'Family History', 'Diagnosis'],
                 data=[
                       ['HG_12_34', '12/3/12', '23', 'Y', 'Jerusalem Syndrome'],
                       ['LG_3_45', '3/4/12', '45', 'N', 'Paris Syndrome'],
                       ['HG_12_34_2', '4/5/13', '-', '-', '-'],
                       ['KD_89_9', '8/9/12', '-', '-', '-'],
                       ['KD_98_9_2', '6/1/13', '54', 'Y', 'Chronic Hiccups'],
                       ['LG_3_45_2', '4/4/10', '59', 'N', 'Dangerous Sneezing Syndrome']
                       ])

desired_df = pd.DataFrame(columns=['Sample', 'Date','Age', 'Family History', 'Diagnosis'],
                 data=[
                       ['HG_12_34', '12/3/12', '23', 'Y', 'Jerusalem Syndrome'],
                       ['LG_3_45', '3/4/12', '45', 'N', 'Paris Syndrome'],
                       ['HG_12_34_2', '4/5/13', '23', 'Y', 'Jerusalem Syndrome'],
                       ['KD_89_9', '8/9/12', '54', 'Y', 'Chronic Hiccups'],
                       ['KD_98_9_2', '6/1/13', '54', 'Y', 'Chronic Hiccups'],
                       ['LG_3_45_2', '4/4/10', '59', 'N', 'Dangerous Sneezing Syndrome']
                       ])

下面详细介绍了我在这方面真正低效和不完整的尝试：

def testing(duplicate, df):
    ''' Checking difference in phenotype data between duplicates
        and return the sample name if 
    '''
    # only assess the duplicate
    if duplicate['Sample'][:-2] in list(df['Sample'].unique()):

        # get sam row
        sam = df[df['Sample'] == duplicate['Sample'][:-2]]

        # store the Age, Family History and Diagnosis in a list for each sample
        sam_pheno = sam.iloc[0][2:4].fillna("-").tolist()
        duplicate_pheno = duplicate[2:4].fillna("-").tolist()

        # if the duplicate sample has nothing in these fields then return the 
        # orginal sample name
        if len(set(duplicate_pheno)) == 1 and list(set(duplicate_pheno))[0] == "-" \
          and len(set(sam_pheno)) > 1:
            return duplicate['Sample'][:-2]         






# this creates a column called Pheno which has the name of the sample which contains the phenotype data that they should share. This is intended so that I can somehow copy over the phenotype data from the sample name in the Pheno field. However, I have no idea how to do this.
input_df['Pheno'] = input_df.apply(lambda x: testing(x, input_df), axis =1)

Answer 1

与耶斯雷尔的回答相似，我想我还是把它留在这里吧，因为我已经写过了。你知道吗

input_df = input_df.replace('-', np.nan)

filter = (input_df["Sample"].str.endswith('_2')) & (input_df["Sample"].str.len() > 7)

samples1 = input_df[~filter].copy()
samples2 = input_df[filter].copy()

samples2['Sample'] = samples2["Sample"].str.replace("_2","")

samples1 = samples1.set_index("Sample")
samples2 = samples2.set_index("Sample")

samples1 = samples1.combine_first(samples2)
samples2 = samples2.combine_first(samples1)

samples1 = samples1.reset_index()
samples2 = samples2.reset_index()

samples2["Sample"] = samples2["Sample"] + "_2"

samples = pd.concat([samples1,samples2])

Answer 2

您可以使用：

#replace all - values to NaN
input_df = input_df.replace('-',np.nan)
#all values end with _2 and longer as 7
mask = (input_df.Sample.str.endswith('_2')) & (input_df.Sample.str.len() > 7)
#create new columnn same with column Sample + remove last 2 chars (_2)
input_df.ix[mask, 'same'] = input_df.ix[mask, 'Sample'].str[:-2]
#replace NaN in same by Sample column
input_df.same = input_df.same.combine_first(input_df.Sample)
#sort values
input_df = input_df.sort_values(['same','Family History'], ascending=False)
#replace NaN by forward filling
input_df[['Age','Family History','Diagnosis']] = 
input_df[['Age','Family History','Diagnosis']].ffill()
#get original index by sorting
input_df.sort_index(inplace=True)
#remove column same
input_df.drop('same', axis=1, inplace=True)

print (input_df)     
       Sample     Date Age Family History                    Diagnosis
0    HG_12_34  12/3/12  23              Y           Jerusalem Syndrome
1     LG_3_45   3/4/12  45              N               Paris Syndrome
2  HG_12_34_2   4/5/13  23              Y           Jerusalem Syndrome
3     KD_89_9   8/9/12  54              Y              Chronic Hiccups
4   KD_98_9_2   6/1/13  54              Y              Chronic Hiccups
5   LG_3_45_2   4/4/10  59              N  Dangerous Sneezing Syndrome

---

print (desired_df)                   
       Sample     Date Age Family History                    Diagnosis
0    HG_12_34  12/3/12  23              Y           Jerusalem Syndrome
1     LG_3_45   3/4/12  45              N               Paris Syndrome
2  HG_12_34_2   4/5/13  23              Y           Jerusalem Syndrome
3     KD_89_9   8/9/12  54              Y              Chronic Hiccups
4   KD_98_9_2   6/1/13  54              Y              Chronic Hiccups
5   LG_3_45_2   4/4/10  59              N  Dangerous Sneezing Syndrome