解决方案1：

Question

我正在处理一个数据框，该数据框包含一列，其数字格式为： [[45, 45, 'D'],[46, 49, 'C'],[50, 66, 'S'],[67, 101, 'C'],[102, 103, 'S'],[104, 106, 'C'],[107, 108, 'S'],[109, 120, 'C'],[121, 121, 'S'],[122, 123, 'C'],[124, 140, 'S'],[141, 149, 'C'],[150, 176, 'S'],[177, 178, 'C'],[179, 181, 'S'],[182, 194, 'C'],[195, 213, 'S'],[214, 21``7, 'C']]

这些数字与字符串中字符的位置相对应：即字符串： 'MGILSFLPVLATESDWADCKSPQPWGHMLLWTAVLFLAPVAGTPAAPPKAVLKLEPQWINVLQEDSVTLTCRGTHSPESDSIQWFHNGNLIPTHTQPSYRFKANNNDSGEYTCQTGQTSLSDPVHLTVLSEWLVLQTPHLEFQEGETIVLRCHSWKDKPLVKVTFFQNGKSKKFSRSDPNFSIPQANHSHSGDYHCTGNIGYTLYSSKPVTITVQAPSSSPMGIIVAVVTGIAVAAIVAAVVALIYCRKKRISALPGYPECREMGETLPEKPANPTNPDEADKVGAENTITYSLLMHPDALEEPDDQNRI'

如您所见，列表中的某些字符与数字列表中的数字不对应（例如，缺少0-44）。因此，必须删除第0-44位的字符以创建较短的字母序列。

我能够对一行执行此操作，但是我正在努力对数据帧中的每一行执行此操作。

这是用于一行的代码：

new_s = ''

for item in res:
    new_s += strSeq[item[0]-1:item[1]]

print(len(new_s), new_s)

这就是我一直试图在所有行中获取的内容：

shortenedSeq_list =[] 
counter=0
stringstring=[]
for rows in df.itertuples():
    strSeq2 = [rows.sequence]
    strremove2 = [rows.shortened_mobidb_consensus]
    for item in strremove2:
        res = ast.literal_eval(item)
    for item in res:
        stringstring.append(strSeq2[item[0]-1:item[1]])

stringstring

但这会导致输出：

 [],
 [],
 [],
 [],
 [],
 [],
 [],
 [],
 [],
 ['MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGSSHAKGIVLEKIGIESKQPNSAIRKCVRVQLIKNGKKVTAFVPNDGCLNFVDENDEVLLAGFGRKGKAKGDIPGVRFKVVKVSGVSLLALWKEKKEKPRS'],
 [],
 [],

鉴于此，我希望列表中的每一行都是缩短的序列。

我最终希望将此列表添加为数据框中的一列。

更新

数字是作为字符串而不是列表输出的，因此res是作为列表的数字，这是工作代码输出：

173 AAPPKAVLKLEPQWINVLQEDSVTLTCRGTHSPESDSIQWFHNGNLIPTHTQPSYRFKANNNDSGEYTCQTGQTSLSDPVHLTVLSEWLVLQTPHLEFQEGETIVLRCHSWKDKPLVKVTFFQNGKSKKFSRSDPNFSIPQANHSHSGDYHCTGNIGYTLYSSKPVTITVQAP，其中173是缩短序列的长度，后跟序列。

df示例：

shortened_mobidb_consensus;sequence
[[45, 45, 'D'], [46, 49, 'C'], [50, 66, 'S'], [67, 101, 'C'], [102, 103, 'S'], [104, 106, 'C'], [107, 108, 'S'], [109, 120, 'C'], [121, 121, 'S'], [122, 123, 'C'], [124, 140, 'S'], [141, 149, 'C'], [150, 176, 'S'], [177, 178, 'C'], [179, 181, 'S'], [182, 194, 'C'], [195, 213, 'S'], [214, 217, 'C']];MGILSFLPVLATESDWADCKSPQPWGHMLLWTAVLFLAPVAGTPAAPPKAVLKLEPQWINVLQEDSVTLTCRGTHSPESDSIQWFHNGNLIPTHTQPSYRFKANNNDSGEYTCQTGQTSLSDPVHLTVLSEWLVLQTPHLEFQEGETIVLRCHSWKDKPLVKVTFFQNGKSKKFSRSDPNFSIPQANHSHSGDYHCTGNIGYTLYSSKPVTITVQAPSSSPMGIIVAVVTGIAVAAIVAAVVALIYCRKKRISALPGYPECREMGETLPEKPANPTNPDEADKVGAENTITYSLLMHPDALEEPDDQNRI
[[1, 1, 'D'], [2, 143, 'S'], [144, 145, 'C']];MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGSSHAKGIVLEKIGIESKQPNSAIRKCVRVQLIKNGKKVTAFVPNDGCLNFVDENDEVLLAGFGRKGKAKGDIPGVRFKVVKVSGVSLLALWKEKKEKPRS
[[1, 145, 'S']];MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGSSHAKGIVLEKIGIESKQPNSAIRKCVRVQLIKNGKKVTAFVPNDGCLNFVDENDEVLLAGFGRKGKAKGDIPGVRFKVVKVSGVSLLALWKEKKEKPRS
[[1, 1, 'D'], [2, 2, 'C'], [3, 37, 'S'], [38, 39, 'C'], [40, 40, 'S'], [41, 41, 'C'], [42, 62, 'S'], [63, 65, 'C'], [66, 231, 'S']];MSKNILVLGGSGALGAEVVKFFKSKSWNTISIDFRENPNADHSFTIKDSGEEEIKSVIEKINSKSIKVDTFVCAAGGWSGGNASSDEFLKSVKGMIDMNLYSAFASAHIGAKLLNQGGLFVLTGASAALNRTSGMIAYGATKAATHHIIKDLASENGGLPAGSTSLGILPVTLDTPTNRKYMSDANFDDWTPLSEVAEKLFEWSTNSDSRPTNGSLVKFETKSKVTTWTNL
[[24, 29, 'D'], [30, 91, 'S'], [92, 92, 'D']];MKVSTTALAVLLCTMTLCNQVFSAPYGADTPTACCFSYSRKIPRQFIVDYFETSSLCSQPGVIFLTKRNRQICADSKETWVQEYITDLELNA

Answer 1

解决方案1：

df = pd.read_csv('stringsample.txt',sep=';',converters={0:ast.literal_eval})

for index, row in df.iterrows():
    new_s = ''
    res = row.shortened_mobidb_consensus
    for item in res:
        new_s += row.sequence[item[0]-1:item[1]]
    df.loc[index,'output'] = new_s

df['output']

0    AAPPKAVLKLEPQWINVLQEDSVTLTCRGTHSPESDSIQWFHNGNL...
1    MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGS...
2    MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGS...
3    MSKNILVLGGSGALGAEVVKFFKSKSWNTISIDFRENPNADHSFTI...
4    APYGADTPTACCFSYSRKIPRQFIVDYFETSSLCSQPGVIFLTKRN...
Name: output, dtype: object

解决方案2 ：（修复您的代码）

df = pd.read_csv('stringsample.txt',sep=';')
shortenedSeq_list =[] 
counter=0
stringstring=[]
for rows in df.itertuples():
    strSeq2 = rows.sequence
    strremove2 = rows.shortened_mobidb_consensus
    res = ast.literal_eval(strremove2)
    new_s = ''
    for item in res:
        new_s += strSeq2[item[0]-1:item[1]]
    stringstring.append(new_s)

stringstring

['AAPPKAVLKLEPQWINVLQEDSVTLTCRGTHSPESDSIQWFHNGNLIPTHTQPSYRFKANNNDSGEYTCQTGQTSLSDPVHLTVLSEWLVLQTPHLEFQEGETIVLRCHSWKDKPLVKVTFFQNGKSKKFSRSDPNFSIPQANHSHSGDYHCTGNIGYTLYSSKPVTITVQAP',
 'MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGSSHAKGIVLEKIGIESKQPNSAIRKCVRVQLIKNGKKVTAFVPNDGCLNFVDENDEVLLAGFGRKGKAKGDIPGVRFKVVKVSGVSLLALWKEKKEKPRS',
 'MGKGKPRGLNSARKLRVHRRNNRWAETTYKKRLLGTAFKSSPFGGSSHAKGIVLEKIGIESKQPNSAIRKCVRVQLIKNGKKVTAFVPNDGCLNFVDENDEVLLAGFGRKGKAKGDIPGVRFKVVKVSGVSLLALWKEKKEKPRS',
 'MSKNILVLGGSGALGAEVVKFFKSKSWNTISIDFRENPNADHSFTIKDSGEEEIKSVIEKINSKSIKVDTFVCAAGGWSGGNASSDEFLKSVKGMIDMNLYSAFASAHIGAKLLNQGGLFVLTGASAALNRTSGMIAYGATKAATHHIIKDLASENGGLPAGSTSLGILPVTLDTPTNRKYMSDANFDDWTPLSEVAEKLFEWSTNSDSRPTNGSLVKFETKSKVTTWTNL',
 'APYGADTPTACCFSYSRKIPRQFIVDYFETSSLCSQPGVIFLTKRNRQICADSKETWVQEYITDLELNA']

迭代添加新字符串作为列表中的元素

1 个答案:

解决方案1：

解决方案2 ：（修复您的代码）