Here's the pandas dataframe that I'm using to learn how to do this:
import pandas as pd
test_list = pd.DataFrame()
test_list["Item"] = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"]
test_list["Number"] = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11"]
test_list["Combined Numbers"]= ""
Based on that dataframe above, I intend to combine up to 3 numbers, separated by commas.
Following that, I intend to repeat this combined value I now have, for each of the test_list["Item"] and test_list["Number"] involved.
I've been scratching my head figuring it out so far. So far I've seen examples of groupby() function for situations like combining information based on a given criteria, like a duplicate value from a column. I'm learning to explore if I don't have anything to refer to, how can I work this out instead?
Here's my intended goal:
Item
Number
Combined Numbers
A
1
1, 2, 3
B
2
1, 2, 3
C
3
1, 2, 3
D
4
4, 5, 6
E
5
4, 5, 6
F
6
4, 5, 6
G
7
7, 8, 9
H
8
7, 8, 9
I
9
7, 8, 9
J
10
10, 11
K
11
10, 11
Thank you
In you case as you don't have this column with the groups, you can generate a range of the length of your dataframe (with np.arange) and do the floor division by 3 (//3). Use groupby.transform to keep the original shape of your data and do the join operation on the column.
test_list["Combined Numbers"] = (
test_list.groupby(np.arange(len(test_list))//3)
['Number'].transform(', '.join)
)
print(test_list)
# Item Number Combined Numbers
# 0 A 1 1, 2, 3
# 1 B 2 1, 2, 3
# 2 C 3 1, 2, 3
# 3 D 4 4, 5, 6
# 4 E 5 4, 5, 6
# 5 F 6 4, 5, 6
# 6 G 7 7, 8, 9
# 7 H 8 7, 8, 9
# 8 I 9 7, 8, 9
# 9 J 10 10, 11
# 10 K 11 10, 11
Related
I need to print the result of groupby object in Python for a specific group/groups only.
Below is the dataframe:
import pandas as pd
df = pd.DataFrame({'ID' : [1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4],
'Entry' : [1, 2, 3, 4, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6]})
print("\n df = \n",df)
In order to group the dataferame by ID and print the result I used these codes:
grouped_by_unit = df.groupby(by="ID")
print("\n", grouped_by_unit.apply(print))
Can somebody please let me know below two things:
How can I print the data frame grouped by 'ID=1' only?
I need to get the below output:
Likewise, how can I print the data frame grouped by 'ID=1' and 'ID=4' together?
I need to get the below output:
You can iterate over the groups for example with for-loop:
grouped_by_unit = df.groupby(by="ID")
for id_, g in grouped_by_unit:
if id_ == 1 or id_ == 4:
print(g)
print()
Prints:
ID Entry
0 1 1
1 1 2
2 1 3
3 1 4
ID Entry
12 4 1
13 4 2
14 4 3
15 4 4
16 4 5
17 4 6
You can use get_group function:
df.groupby(by="ID").get_group(1)
which prints
ID Entry
0 1 1
1 1 2
2 1 3
3 1 4
You can use the same method to print the group for the key 4.
As the title says:
a = pd.DataFrame([1,2,3,4,5,6,7,8,9,10])
Having a dataframe with 10 values we want to aggregate say last 5 rows and put them as list into a new column:
>>> a new_col
0
0 1
1 2
2 3
3 4
4 5 [1,2,3,4,5]
5 6 [2,3,4,5,6]
6 7 [3,4,5,6,7]
7 8 [4,5,6,7,8]
8 9 [5,6,7,8,9]
9 10 [6,7,8,9,10]
How?
Due to how rolling windows are implemented, you won't be able to aggregate the results as you expect, but we still can reach your desired result by iterating each window and storing the values as a list of values:
>>> new_col_values = [
window.to_list() if len(window) == 5 else None
for window in df["column"].rolling(5)
]
>>> df["new_col"] = new_col_values
>>> df
column new_col
0 1 None
1 2 None
2 3 None
3 4 None
4 5 [1, 2, 3, 4, 5]
5 6 [2, 3, 4, 5, 6]
6 7 [3, 4, 5, 6, 7]
7 8 [4, 5, 6, 7, 8]
8 9 [5, 6, 7, 8, 9]
9 10 [6, 7, 8, 9, 10]
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 have been having troubles with extracting reading/manipulating/extracting data from a txt file. In the text file it has a general header with various information that is setup something like this below just as an example:
~ECOLOGY
~LOCATION
LAT: 59
LONG: 23
~PARAMETERS
Area. 8
Distribution. 3
Diversity. 5
~DATA X Y CONF DECID PEREN
3 6 1 3 0
7 2 4 2 1
4 8 0 6 2
9 9 6 2 0
2 3 2 5 4
6 5 0 2 7
7 1 2 4 2
I want to be able to extract the headers of the columns and use the headers of the columns as an index or key since sometimes the types of column data can change between files and the amount of rows of data can fluctuate as well. I want to be able to read the data in each column so that pending on location I can sum or add columns such as show below and export it as a separate file:
~DATA X Y CONF DECID PEREN TOTAL
3 6 1 3 0 4
7 2 4 2 1 7
4 8 0 6 2 8
9 9 6 2 0 8
2 3 2 5 4 11
6 5 0 2 7 9
7 1 2 4 2 8
Any suggestions?
This is what I have so far:
E = open("ECOLOGY.txt", "r")
with open(path) as E:
for i, line in enumerate(E):
sep_lines = line.rsplit()
if "~DATA" in sep_lines:
key =(line.rsplit())
key.remove('~DATA')
for j, value in enumerate(key):
print (j,value)
print (key)
dict = {L: v for v, L in enumerate(key)}
print(dict)
Life would be much easier for you if you learned a smidgen of Pandas. But you can do it without.
with open('ttl.txt') as ttl:
for _ in range(10):
next(ttl)
first = True
for line in ttl:
line = line.rstrip()
if first:
first = False
labels = line.split()+['TOTAL']
fmt = 7*'{:<9s}'
print (fmt.format(*labels))
else:
numbers = [int(_) for _ in line.split()]
total = sum(numbers[-3:])
other_items = numbers + [total]
fmt = 6*'{:<9d}'
fmt = '{:<9s}'+fmt
print (fmt.format('', *other_items))
~DATA X Y CONF DECID PEREN TOTAL
3 6 1 3 0 4
7 2 4 2 1 7
4 8 0 6 2 8
9 9 6 2 0 8
2 3 2 5 4 11
6 5 0 2 7 9
7 1 2 4 2 8
next skips lines in the input file. You can use split() to split input lines on whitespace, the use formatting to put items back together as you want them.
This a very basic, frail, format depending solution. But I hope it can help you.
with open("test.txt") as f:
data_part_reached = False
for line in f:
if "~DATA" in line:
column = [[elem] for elem in line.split(" ") if elem not in (" ", "", "\n", "~DATA")]
data_part_reached = True
elif data_part_reached:
values = [int(elem) for elem in line.split(" ") if elem not in (" ", "", "\n")]
for i in range(len(columns)):
columns[i].append(values[i])
columns =
[['X', 3, 7, 4, 9, 2, 6, 7],
['Y', 6, 2, 8, 9, 3, 5, 1],
['CONF', 1, 4, 0, 6, 2, 0, 2],
['DECID', 3, 2, 6, 2, 5, 2, 4],
['PEREN', 0, 1, 2, 0, 4, 7, 2],
['TOTAL', 4, 7, 8, 8, 11, 9, 8]]
This will get you a list of lists where the first element of each list is the header and the rest are the values. I casted the values to int since you said you want to operate with them. You can turn this list into a dict where the key is the header and the list of values of each column are the value if you want, like this.
d = {}
for column in columns:
d[column.pop(0)] = column
d =
{'DECID': [3, 2, 6, 2, 5, 2, 4],
'PEREN': [0, 1, 2, 0, 4, 7, 2],
'CONF': [1, 4, 0, 6, 2, 0, 2],
'X': [3, 7, 4, 9, 2, 6, 7],
'TOTAL': [4, 7, 8, 8, 11, 9, 8],
'Y': [6, 2, 8, 9, 3, 5, 1]}
Create a empty dictionary to store all needed data.
Read from the file object as E and loop until you reach a line starting with ~DATA.
Then split the header items, append TOTAL and then break from the loop.
Create a list to store the remaining data.
Loop to split the data and then append the sum total.
The list will append each list of data.
Loop ends and then adds to list of lists to the dictionary.
dic = {}
with open("ECOLOGY.txt") as E:
for line in E:
if line[:5] == '~DATA':
dic['header'] = line.split()[1:] + ['TOTAL']
break
data = []
for line in E:
cols = line.split()
cols.append(sum([int(num) for num in cols[2:]]))
data.append(cols)
dic['data'] = data
The dictionary will be i.e. {'header': [...], 'data': [[...], ...]}
edit: Added missing dic declaration at the beginning of code.
This question already has an answer here:
Numpy intersect1d with array with matrix as elements
(1 answer)
Closed 5 years ago.
I'm currently trying to compare two matrices and return matching rows into the "intersection matrix" via python. Both matrices are numerical data-and I'm trying to return the rows of their common entries (I have also tried just creating a matrix with matching positional entries along the first column and then creating an accompanying tuple). These matrices are not necessarily the same in dimensionality.
Let's say I have two matrices of matching column length but arbitrary (can be very large and different row length)
23 3 4 5 23 3 4 5
12 6 7 8 45 7 8 9
45 7 8 9 34 5 6 7
67 4 5 6 3 5 6 7
I'd like to create a matrix with the "intersection" being for this low dimensional example
23 3 4 5
45 7 8 9
perhaps it looks like this though:
1 2 3 4 2 4 6 7
2 4 6 7 4 10 6 9
4 6 7 8 5 6 7 8
5 6 7 8
in which case we only want:
2 4 6 7
5 6 7 8
I've tried things of this nature:
def compare(x):
# This is a matrix I created with another function-purely numerical data of arbitrary size with fixed column length D
y =n_c(data_cleaner(x))
# this is a second matrix that i'd like to compare it to. note that the sizes are probably not the same, but the columns length are
z=data_cleaner(x)
# I initialized an array that would hold the matching values
compare=[]
# create nested for loop that will check a single index in one matrix over all entries in the second matrix over iteration
for i in range(len(y)):
for j in range(len(z)):
if y[0][i] == z[0][i]:
# I want the row or the n tuple (shown here) of those columns with the matching first indexes as shown above
c_vec = ([0][i],[15][i],[24][i],[0][25],[0][26])
compare.append(c_vec)
else:
pass
return compare
compare(c_i_w)
Sadly, I'm running into some errors. Specifically it seems that I'm telling python to improperly reference values.
Consider the arrays a and b
a = np.array([
[23, 3, 4, 5],
[12, 6, 7, 8],
[45, 7, 8, 9],
[67, 4, 5, 6]
])
b = np.array([
[23, 3, 4, 5],
[45, 7, 8, 9],
[34, 5, 6, 7],
[ 3, 5, 6, 7]
])
print(a)
[[23 3 4 5]
[12 6 7 8]
[45 7 8 9]
[67 4 5 6]]
print(b)
[[23 3 4 5]
[45 7 8 9]
[34 5 6 7]
[ 3 5 6 7]]
Then we can broadcast and get an array of equal rows with
x = (a[:, None] == b).all(-1)
print(x)
[[ True False False False]
[False False False False]
[False True False False]
[False False False False]]
Using np.where we can identify the indices
i, j = np.where(x)
Show which rows of a
print(a[i])
[[23 3 4 5]
[45 7 8 9]]
And which rows of b
print(b[j])
[[23 3 4 5]
[45 7 8 9]]
They are the same! That's good. That's what we wanted.
We can put the results into a pandas dataframe with a MultiIndex with row number from a in the first level and row number from b in the second level.
pd.DataFrame(a[i], [i, j])
0 1 2 3
0 0 23 3 4 5
2 1 45 7 8 9