小编chi*_*tra的帖子

我有一个数据框,如:

g1   g2   g3   g4   g5
2    0    1    0    1
2    1    1    0    1
2    1    1    2    1

我想删除每个列中至少有一个2的列.

并得到一个新的df:

g2   g3   g5
0    1    1
1    1    1
1    1    1

谢谢你的帮助.

我有一个数据框，例如：

query   qstart  qend    name    number  strand
A       2.0     1064.0  None    0       +
B       2.0     1076.0  None    0       +
C       2.0     1064.0  None    0       +
D       0.0     741.0   None    0       +

我想删除所有小数并得到：

query   qstart  qend    name    number  strand
A       2       1064    None    0       +
B       2       1064    None    0       +
C       2       1064    None    0       +
D       0       741     None    0       +

我试过：

df['qstart']= round(df['qstart'])
df['qend']= round(df['qend'])

但它不起作用......

python中是否有一种简单的方法可以用另一个替换倍数字符？

例如，我想更改：

name1_22:3-3(+):Pos_bos 

至

name1_22_3-3_+__Pos_bos

因此，基本上全部替换"(",")",":"为"_"。

我只知道这样做：

str.replace(":","_")
str.replace(")","_")
str.replace("(","_")

I have a list such as:

list=["Chrm_23-56_python_regius","Chrm_3-89_elephant_regius",
      "Chrm_13-56_monkey_regius","Chrm_13-34_rat_regius","Chrm_67-123_python_regius",
      "chrm_90-345_elephant_regius","Chrm_67-124_monkey_regius",
      "Chrm_345-456_rat_regius","Chrm_789-1000_python_regius"]

and the idea is to put all element in a dict form that have the same name (without the number-number) part. and get something like:

dict = {'key1': ['Chrm_23-56_python_regius','Chrm_67-123_python_regius','Chrm_789-1000_python_regius'],
        'key2': ['Chrm_3-89_elephant_regius','chrm_90-345_elephant_regius'],
        'key3': ['Chrm_13-56_monkey_regius','Chrm_67-124_monkey_regius'],
        'key4': ['Chrm_13-34_rat_regius','Chrm_345-456_rat_regius']}

As you can see for instance, in the key1, the 3 values (without the number-number) are = to Chrm__python_regius.

I know how to see which element is the same without the number-number …