根据字符串内的浮点数对列值进行排序，然后进行连接

Question

我正在处理一个相当混乱的DF。看起来像这样，但是有30列:

<表类> b some text (other text): 56.3% (text again: 40%) again text (not same text): 33% (text text: 60.1%) text (always text): 26.6% (and text: 80%) still text (too text:太多文本):86% (last text: 10%)

您可以尝试apply一个自定义函数

def concat(row):
keys = row.str.extract('(d+.?d*)%')[0].astype(float).tolist()
row = [x for _, x in sorted(zip(keys, row.tolist()))]
return ' '.join(row)
df['c'] = df.apply(concat, axis=1)

print(df)
a                                                  b
0  some text (other text) : 56.3% (text again: 40%)  again text (not same text) : 33% (text text: 6...
1      text (always text) : 26.6% (aaand text: 80%)  still text (too much text) : 86% (last text: 10%)
a  
0  some text (other text) : 56.3% (text again: 40%)
1      text (always text) : 26.6% (aaand text: 80%)
b  
0  again text (not same text) : 33% (text text: 60.1%)
1    still text (too much text) : 86% (last text: 10%)
                                c
0  again text (not same text) : 33% (text text: 60.1%) some text (other text) : 56.3% (text again: 40%)
1        text (always text) : 26.6% (aaand text: 80%) still text (too much text) : 86% (last text: 10%)