Slicing multiple column ranges from a dataframe using iloc

Question

I have a df with 32 columns

df.shape
(568285, 32)

I am trying to rearrange the columns in a specific way, and drop the first column using

Answer 1

You could use the np.r_ indexer.

class RClass(AxisConcatenator)
 |  Translates slice objects to concatenation along the first axis.
 |  
 |  This is a simple way to build up arrays quickly. There are two use cases.

---

df = df.iloc[:, np.r_[31, 1:23, 24, 25, 26, 28, 27, 29, 30]]

---

df

     0     1     2     3     4     5     6     7     8     9   ...     40  \
A  33.0  44.0  68.0  31.0   NaN  87.0  66.0   NaN  72.0  33.0  ...   71.0   
B   NaN   NaN  77.0  98.0   NaN  48.0  91.0  43.0   NaN  89.0  ...   38.0   
C  45.0  55.0   NaN  72.0  61.0  87.0   NaN  99.0  96.0  75.0  ...   83.0   
D   NaN   NaN   NaN  58.0   NaN  97.0  64.0  49.0  52.0  45.0  ...   63.0   

     41    42    43    44    45    46    47    48    49  
A   NaN  87.0  31.0  50.0  48.0  73.0   NaN   NaN  81.0  
B  79.0  47.0  51.0  99.0  59.0   NaN  72.0  48.0   NaN  
C  93.0   NaN  95.0  97.0  52.0  99.0  71.0  53.0  69.0  
D   NaN  41.0   NaN   NaN  55.0  90.0   NaN   NaN  92.0

out = df.iloc[:, np.r_[31, 1:23, 24, 25, 26, 28, 27, 29, 30]]
out 
     31    1     2     3     4     5     6     7     8     9   ...     20  \
A  99.0  44.0  68.0  31.0   NaN  87.0  66.0   NaN  72.0  33.0  ...   66.0   
B  42.0   NaN  77.0  98.0   NaN  48.0  91.0  43.0   NaN  89.0  ...    NaN   
C  77.0  55.0   NaN  72.0  61.0  87.0   NaN  99.0  96.0  75.0  ...   76.0   
D  95.0   NaN   NaN  58.0   NaN  97.0  64.0  49.0  52.0  45.0  ...   71.0   

     21    22    24    25    26    28    27    29    30  
A   NaN  40.0  66.0  87.0  97.0  68.0   NaN  68.0   NaN  
B  95.0   NaN  47.0  79.0  47.0   NaN  83.0  81.0  57.0  
C   NaN  75.0  46.0  84.0   NaN  50.0  41.0  38.0  52.0  
D   NaN  74.0  41.0  55.0  60.0   NaN   NaN  84.0   NaN