I have a large dataset of 670 columns and 2856 rows. The idea is to sum two consequent rows and retrieve a single column and value as result. It’s important to not have replacement in the way the fist column + second, then the third + the fourth **not** the second + third.

Index | ID1 | ID2 | ID3 | ID4 |
---|---|---|---|---|

First | 0 | 1 | 0 | 1 |

Second | 0 | 0 | 1 | 1 |

the result should be

Index | ID12 | ID34 |
---|---|---|

First | 1 | 1 |

Second | 0 | 2 |

The example dataframe:

```
df = pd.DataFrame({"ID1" : [0,0,0,1,1,1] , "ID2" :[1,1,1,0,0,0], "ID3" : [0,1,1,1,0,1]},"ID4" : [0,0,0,0,0,0])
result = pd.DataFrame({"ID1/2" : [1,1,1,0,0,0] , "ID3/4" :[0,1,1,1,0,1]})
```

I have tried:

```
res = []
for i in range(len(df)):
for j in range(1,len(df.columns),2):
res.append(data.iloc[i,j]+data.iloc[i,j-1])
result = pd.DataFrame(res)
```

In **R** the result is:

```
result <- matrix(nrow = nrow(df), ncol = ncol(df),)
for (i in seq(1,ncol(df),2)){
result[,i] <- df[,i]+df[,i+1]
}
#Erasing the NAs columns
result <- result [,-seq(2,ncol(result ),2)]
```

### >Solution :

```
N = len(df.columns)
new_df = df.groupby(np.arange(N) // 2, axis="columns").sum()
new_df.columns = [f"ID{j}{j+1}" for j in range(1, N, 2)]
```

- groupby every-2 columns and sum
- floordivision of the range(len(columns)) by 2 gives the group numbers: 0, 0, 1, 1, 2, 2…

- form new columns with stepping by 2 over range(N) to get 1, 3…

to get

```
>>> new_df
ID12 ID34
Index
First 1 1
Second 0 2
```