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I have a very large dataframe with 1,000 columns. The first few columns occur only once, denoting a customer. The next few columns are representative of multiple encounters with the customer, with an underscore and the number encounter. Every additional encounter adds a new column, so there is NOT a fixed number of columns -- it'll grow with time.
Sample dataframe header structure excerpt:
id dob gender pro_1 pro_10 pro_11 pro_2 ... pro_9 pre_1 pre_10 ...
I'm trying to re-order the columns based on the number after the column name, so all _1 should be together, all _2 should be together, etc, like so:
id dob gender pro_1 pre_1 que_1 fre_1 gen_1 pro2 pre_2 que_2 fre_2 ...
(Note that the re-order should order the numbers correctly; the current order treats them like strings, which orders 1, 10, 11, etc. rather than 1, 2, 3)
Is this possible to do in pandas, or should I be looking at something else? Any help would be greatly appreciated! Thank you!
EDIT:
Alternatively, is it also possible to re-arrange column names based on the string part AND number part of the column names? So the output would then look similar to the original, except the numbers would be considered so that the order is more intuitive:
id dob gender pro_1 pro_2 pro_3 ... pre_1 pre_2 pre_3 ...
EDIT 2.0:
Just wanted to thank everyone for helping! While only one of the responses worked, I really appreciate the effort and learned a lot about other approaches / ways to think about this.
Here is one way you can try:
# column names copied from your example
example_cols = 'id dob gender pro_1 pro_10 pro_11 pro_2 pro_9 pre_1 pre_10'.split()
# sample DF
df = pd.DataFrame([range(len(example_cols))], columns=example_cols)
df
# id dob gender pro_1 pro_10 pro_11 pro_2 pro_9 pre_1 pre_10
#0 0 1 2 3 4 5 6 7 8 9
# number of columns excluded from sorting
N = 3
# get a list of columns from the dataframe
cols = df.columns.tolist()
# split, create an tuple of (column_name, prefix, number) and sorted based on the 2nd and 3rd item of the tuple, then retrieved the first item.
# adjust "key = lambda x: x[2]" to group cols by numbers only
cols_new = cols[:N] + [ a[0] for a in sorted([ (c, p, int(n)) for c in cols[N:] for p,n in [c.split('_')]], key = lambda x: (x[1], x[2])) ]
# get the new dataframe based on the cols_new
df_new = df[cols_new]
# id dob gender pre_1 pre_10 pro_1 pro_2 pro_9 pro_10 pro_11
#0 0 1 2 8 9 3 6 7 4 5
Luckily there is a one liner in python that can fix this:
df = df.reindex(sorted(df.columns), axis=1)
For Example lets say you had this dataframe:
import pandas as pd
import numpy as np
df = pd.DataFrame({'Name': [2, 4, 8, 0],
'ID': [2, 0, 0, 0],
'Prod3': [10, 2, 1, 8],
'Prod1': [2, 4, 8, 0],
'Prod_1': [2, 4, 8, 0],
'Pre7': [2, 0, 0, 0],
'Pre2': [10, 2, 1, 8],
'Pre_2': [10, 2, 1, 8],
'Pre_9': [10, 2, 1, 8]}
)
print(df)
Output:
Name ID Prod3 Prod1 Prod_1 Pre7 Pre2 Pre_2 Pre_9
0 2 2 10 2 2 2 10 10 10
1 4 0 2 4 4 0 2 2 2
2 8 0 1 8 8 0 1 1 1
3 0 0 8 0 0 0 8 8 8
Then used
df = df.reindex(sorted(df.columns), axis=1)
Then the dataframe will then look like:
ID Name Pre2 Pre7 Pre_2 Pre_9 Prod1 Prod3 Prod_1
0 2 2 10 2 10 10 2 10 2
1 0 4 2 0 2 2 4 2 4
2 0 8 1 0 1 1 8 1 8
3 0 0 8 0 8 8 0 8 0
As you can see, the columns without underscore will come first, followed by an ordering based on the number after the underscore. However this also sorts of the column names, so the column names that come first in the alphabet will be first.
You need to split you column on '_' then convert to int:
c = ['A_1','A_10','A_2','A_3','B_1','B_10','B_2','B_3']
df = pd.DataFrame(np.random.randint(0,100,(2,8)), columns = c)
df.reindex(sorted(df.columns, key = lambda x: int(x.split('_')[1])), axis=1)
Output:
A_1 B_1 A_2 B_2 A_3 B_3 A_10 B_10
0 68 11 59 69 37 68 76 17
1 19 37 52 54 23 93 85 3
Next case, you need human sorting:
import re
def atoi(text):
return int(text) if text.isdigit() else text
def natural_keys(text):
'''
alist.sort(key=natural_keys) sorts in human order
http://nedbatchelder.com/blog/200712/human_sorting.html
(See Toothy's implementation in the comments)
'''
return [ atoi(c) for c in re.split(r'(\d+)', text) ]
df.reindex(sorted(df.columns, key = lambda x:natural_keys(x)), axis=1)
Output:
A_1 A_2 A_3 A_10 B_1 B_2 B_3 B_10
0 68 59 37 76 11 69 68 17
1 19 52 23 85 37 54 93 3
Try this.
To re-order the columns based on the number after the column name
cols_fixed = df.columns[:3] # change index no based on your df
cols_variable = df.columns[3:] # change index no based on your df
cols_variable = sorted(cols_variable, key=lambda x : int(x.split('_')[1])) # split based on the number after '_'
cols_new = cols_fixed + cols_variable
new_df = pd.DataFrame(df[cols_new])
To re-arrange column names based on the string part AND number part of the column names
cols_fixed = df.columns[:3] # change index no based on your df
cols_variable = df.columns[3:] # change index no based on your df
cols_variable = sorted(cols_variable)
cols_new = cols_fixed + cols_variable
new_df = pd.DataFrame(df[cols_new])
I feel like there is a better way than this:
import pandas as pd
df = pd.DataFrame(
columns=" index c1 c2 v1 ".split(),
data= [
[ 0, "A", "X", 3, ],
[ 1, "A", "X", 5, ],
[ 2, "A", "Y", 7, ],
[ 3, "A", "Y", 1, ],
[ 4, "B", "X", 3, ],
[ 5, "B", "X", 1, ],
[ 6, "B", "X", 3, ],
[ 7, "B", "Y", 1, ],
[ 8, "C", "X", 7, ],
[ 9, "C", "Y", 4, ],
[ 10, "C", "Y", 1, ],
[ 11, "C", "Y", 6, ],]).set_index("index", drop=True)
def callback(x):
x['seq'] = range(1, x.shape[0] + 1)
return x
df = df.groupby(['c1', 'c2']).apply(callback)
print df
To achieve this:
c1 c2 v1 seq
0 A X 3 1
1 A X 5 2
2 A Y 7 1
3 A Y 1 2
4 B X 3 1
5 B X 1 2
6 B X 3 3
7 B Y 1 1
8 C X 7 1
9 C Y 4 1
10 C Y 1 2
11 C Y 6 3
Is there a way to do it that avoids the callback?
use cumcount(), see docs here
In [4]: df.groupby(['c1', 'c2']).cumcount()
Out[4]:
0 0
1 1
2 0
3 1
4 0
5 1
6 2
7 0
8 0
9 0
10 1
11 2
dtype: int64
If you want orderings starting at 1
In [5]: df.groupby(['c1', 'c2']).cumcount()+1
Out[5]:
0 1
1 2
2 1
3 2
4 1
5 2
6 3
7 1
8 1
9 1
10 2
11 3
dtype: int64
This might be useful
df = df.sort_values(['userID', 'date'])
grp = df.groupby('userID')['ItemID'].aggregate(lambda x: '->'.join(tuple(x))).reset_index()
print(grp)
it will create a sequence like this
If you have a dataframe similar to the one below and you want to add seq column by building it from c1 or c2, i.e. keep a running count of similar values (or until a flag comes up) in other column(s), read on.
df = pd.DataFrame(
columns=" c1 c2 seq".split(),
data= [
[ "A", 1, 1 ],
[ "A1", 0, 2 ],
[ "A11", 0, 3 ],
[ "A111", 0, 4 ],
[ "B", 1, 1 ],
[ "B1", 0, 2 ],
[ "B111", 0, 3 ],
[ "C", 1, 1 ],
[ "C11", 0, 2 ] ])
then first find group starters, (str.contains() (and eq()) is used below but any method that creates a boolean Series such as lt(), ne(), isna() etc. can be used) and call cumsum() on it to create a Series where each group has a unique identifying value. Then use it as the grouper on a groupby().cumsum() operation.
In summary, use a code similar to the one below.
# build a grouper Series for similar values
groups = df['c1'].str.contains("A$|B$|C$").cumsum()
# or build a grouper Series from flags (1s)
groups = df['c2'].eq(1).cumsum()
# groupby using the above grouper
df['seq'] = df.groupby(groups).cumcount().add(1)
The cleanliness of Jeff's answer is nice, but I prefer to sort explicitly...though generally without overwriting my df for these type of use-cases (e.g. Shaina Raza's answer).
So, to create a new column sequenced by 'v1' within each ('c1', 'c2') group:
df["seq"] = df.sort_values(by=['c1','c2','v1']).groupby(['c1','c2']).cumcount()
you can check with:
df.sort_values(by=['c1','c2','seq'])
or, if you want to overwrite the df, then:
df = df.sort_values(by=['c1','c2','seq']).reset_index()
I want to turn
arr = np.array([[[1,2,3],[4,5,6],[7,8,9],[10,11,12]], [[2,2,2],[4,5,6],[7,8,9],[10,11,12]], [[3,3,3],[4,5,6],[7,8,9],[10,11,12]]])
into
arr = np.array([[[1,2,3],[7,8,9],[10,11,12]], [[2,2,2],[7,8,9],[10,11,12]], [[3,3,3],[7,8,9],[10,11,12]]])
Below is the code:
a = 0
b = 0
NewArr = []
while a < 3:
c = arr[a, :, :]
d = arr[a]
print(d)
if c[1, 2] == 6:
c = np.delete(c, [1], axis=0)
a += 1
b += 1
c = np.concatenate((d, c), axis=1)
print(c)
But after deleting the line containing the number 6, I cannot stitch the array together，Can someone help me?
thank you very much for your help.
If you want a more automatic way of processing your input data, here is an answer using numpy functions :
arr[np.newaxis,~np.any(arr==6,axis=2)].reshape((3,-1,3))
np.any(arr==6,axis=2) outputs an array which has True at rows which contain the value 6. We take the inverse of those booleans since we want to remove those rows. The solution is then used as an index selection in arr, with a np.newaxis because the output of np.any had one axis less than the original array.
Finally, the output is reshaped into a 3,x,3 array, where x will depend on the number of rows which were removed (hence the -1 in reshape)
Based on the input / output you provide, a simpler solution would be to just use index selection and slices:
import numpy as np
arr = np.array([[[1,2,3],[4,5,6],[7,8,9],[10,11,12]], [[2,2,2],[4,5,6],[7,8,9],[10,11,12]], [[3,3,3],[4,5,6],[7,8,9],[10,11,12]]])
print("arr=")
print(arr)
expected_result = np.array([[[1,2,3],[7,8,9],[10,11,12]], [[2,2,2],[7,8,9],[10,11,12]], [[3,3,3],[7,8,9],[10,11,12]]])
# select indices 0, 2 and 3 from dimension 2
a = np.copy(arr[:,[0,2,3],:])
print("a=")
print(a)
print(np.array_equal(a, expected_result))
Output:
arr=
[[[ 1 2 3]
[ 4 5 6]
[ 7 8 9]
[10 11 12]]
[[ 2 2 2]
[ 4 5 6]
[ 7 8 9]
[10 11 12]]
[[ 3 3 3]
[ 4 5 6]
[ 7 8 9]
[10 11 12]]]
a=
[[[ 1 2 3]
[ 7 8 9]
[10 11 12]]
[[ 2 2 2]
[ 7 8 9]
[10 11 12]]
[[ 3 3 3]
[ 7 8 9]
[10 11 12]]]
True
I am analyzing a dataset that has an Origin ID (Column A), a Destination ID (Column B), and how many trips have happened between them (Column Count). Now I want to sum the A-B trips with the B-A trips. This sum is the total number of trips between A and B.
Here is how my data looks like (it is not necessarily ordered in the same way):
In [1]: group_station = pd.DataFrame([[1, 2, 100], [2, 1, 200], [4, 6, 5] , [6, 4, 10], [1, 4, 70]], columns=['A', 'B', 'Count'])
Out[2]:
A B Count
0 1 2 100
1 2 1 200
2 4 6 5
3 6 4 10
4 1 4 70
And I want the following output:
A B C
0 1 2 300
1 4 6 15
4 1 4 70
I have tried groupby and setting the index to both variables with no success. Right now I am doing a very inefficient double loop, that is too slow for the size of my dataset.
If it helps this is the code for the double loop (I removed some efficiency modifications to make it more clear):
# group_station is the dataframe
collapsed_group_station = np.zeros(len(group_station), 3))
for i, row in enumerate(group_station.iterrows()):
start_id = row[0][0]
end_id = row[0][1]
count = row[1][0]
for check_row in group_station.iterrows():
check_start_id = check_row[0][0]
check_end_id = check_row[0][1]
check_time = check_row[1][0]
if start_id == check_end_id and end_id == check_start_id:
new_group_station[i][0] = start_id
new_group_station[i][1] = end_id
new_group_station[i][2] = time + check_time
break
I have ideas of how to make this code more efficient, but I wanted to know if there is a way of doing it without looping.
You can using np.sort with groupby.sum()
import numpy as np; import pandas as pd
group_station[['A','B']]=np.sort(group_station[['A','B']],axis=1)
group_station.groupby(['A','B'],as_index=False).Count.sum()
Out[175]:
A B Count
0 1 2 300
1 1 4 70
2 4 6 15
I am working with the confusion matrix. So for each loop I have an array (confusion matrix). As I am doing 10 loops, I end up with 10 arrays. I want to sum all of them.
So I decided that for each loop I am going to store the arrays inside a list --I do not know whether it is better to store them inside an array.
And now I want to add each array which is inside the list.
So If I have:
5 0 0 1 1 0
0 5 0 2 4 0
0 0 5 2 0 5
The sum will be:
6 1 0
2 9 0
2 0 10
This is a picture of my confusion matrices and my list of arrays:
This is my code:
list_cm.sum(axis=0)
Just sum the list:
>>> sum([np.array([[5,0,0],[0,5,0],[0,0,5]]), np.array([[1,1,0],[2,4,0],[2,0,5]])])
array([[ 6, 1, 0],
[ 2, 9, 0],
[ 2, 0, 10]])