group DataFrame by column value using pandas
0
I have the following (pandas) DataFrame:
b c d e
0 100 369 203 314
1 100 228 784 366
2 200 811 664 202
3 200 531 932 575
I want to iterate its rows but I need to know where the value of b changed.
I am looking for a way to get this DF content in groups or something like that and then iterate over the rows of each group (with a nested loop and in this way I can get an indication that the b value did changed):
In the first iteration my new df will be:
b c d e
0 100 369 203 314
1 100 228 784 366
In the second iteration my new df will be:
b c d e
0 200 811 664 202
1 200 531 932 575
python python-2.7 pandas
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
|
show 3 more comments
0
I have the following (pandas) DataFrame:
b c d e
0 100 369 203 314
1 100 228 784 366
2 200 811 664 202
3 200 531 932 575
I want to iterate its rows but I need to know where the value of b changed.
I am looking for a way to get this DF content in groups or something like that and then iterate over the rows of each group (with a nested loop and in this way I can get an indication that the b value did changed):
In the first iteration my new df will be:
b c d e
0 100 369 203 314
1 100 228 784 366
In the second iteration my new df will be:
b c d e
0 200 811 664 202
1 200 531 932 575
python python-2.7 pandas
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
1
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
d=x: y for x, y in df.groupby('b')
– Wen-Ben
Nov 13 '18 at 16:16
d = dict(tuple(df.groupby('b')))?
– jezrael
Nov 13 '18 at 16:17
|
show 3 more comments
0
0
0
I have the following (pandas) DataFrame:
b c d e
0 100 369 203 314
1 100 228 784 366
2 200 811 664 202
3 200 531 932 575
I want to iterate its rows but I need to know where the value of b changed.
I am looking for a way to get this DF content in groups or something like that and then iterate over the rows of each group (with a nested loop and in this way I can get an indication that the b value did changed):
In the first iteration my new df will be:
b c d e
0 100 369 203 314
1 100 228 784 366
In the second iteration my new df will be:
b c d e
0 200 811 664 202
1 200 531 932 575
python python-2.7 pandas
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
I have the following (pandas) DataFrame:
b c d e
0 100 369 203 314
1 100 228 784 366
2 200 811 664 202
3 200 531 932 575
I want to iterate its rows but I need to know where the value of b changed.
I am looking for a way to get this DF content in groups or something like that and then iterate over the rows of each group (with a nested loop and in this way I can get an indication that the b value did changed):
In the first iteration my new df will be:
b c d e
0 100 369 203 314
1 100 228 784 366
In the second iteration my new df will be:
b c d e
0 200 811 664 202
1 200 531 932 575
python python-2.7 pandas
python python-2.7 pandas
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
asked Nov 13 '18 at 16:13
Andy ThomasAndy Thomas
4321625
4321625
1
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
d=x: y for x, y in df.groupby('b')
– Wen-Ben
Nov 13 '18 at 16:16
d = dict(tuple(df.groupby('b')))?
– jezrael
Nov 13 '18 at 16:17
|
show 3 more comments
1
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
d=x: y for x, y in df.groupby('b')
– Wen-Ben
Nov 13 '18 at 16:16
d = dict(tuple(df.groupby('b')))?
– jezrael
Nov 13 '18 at 16:17
1
1
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
d=x: y for x, y in df.groupby('b') – Wen-Ben
Nov 13 '18 at 16:16
d=x: y for x, y in df.groupby('b') – Wen-Ben
Nov 13 '18 at 16:16
d = dict(tuple(df.groupby('b'))) ?– jezrael
Nov 13 '18 at 16:17
d = dict(tuple(df.groupby('b'))) ?– jezrael
Nov 13 '18 at 16:17
|
show 3 more comments
1 Answer
1
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oldest
votes
0
You can use the groupby method:
from pandas import DataFrame
columns = ["b", "c", "d", "e"]
data = [[100, 369, 203, 314], [100, 228, 784, 366], [200, 811, 664, 202], [200, 531, 932, 575]]
df = DataFrame(data=data,columns=columns)
def split_dfs(df, col):
return [group[1] for group in list(df.groupby(col))]
dfs = split_dfs(df, "b")
for df_group in dfs:
print(df_group)
b c d e
0 100 369 203 314
1 100 228 784 366
b c d e
2 200 811 664 202
3 200 531 932 575
answered Nov 13 '18 at 16:34
user2921352user2921352
988
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
add a comment |
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1 Answer
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1 Answer
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active
oldest
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0
You can use the groupby method:
from pandas import DataFrame
columns = ["b", "c", "d", "e"]
data = [[100, 369, 203, 314], [100, 228, 784, 366], [200, 811, 664, 202], [200, 531, 932, 575]]
df = DataFrame(data=data,columns=columns)
def split_dfs(df, col):
return [group[1] for group in list(df.groupby(col))]
dfs = split_dfs(df, "b")
for df_group in dfs:
print(df_group)
b c d e
0 100 369 203 314
1 100 228 784 366
b c d e
2 200 811 664 202
3 200 531 932 575
answered Nov 13 '18 at 16:34
user2921352user2921352
988
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
add a comment |
0
You can use the groupby method:
from pandas import DataFrame
columns = ["b", "c", "d", "e"]
data = [[100, 369, 203, 314], [100, 228, 784, 366], [200, 811, 664, 202], [200, 531, 932, 575]]
df = DataFrame(data=data,columns=columns)
def split_dfs(df, col):
return [group[1] for group in list(df.groupby(col))]
dfs = split_dfs(df, "b")
for df_group in dfs:
print(df_group)
b c d e
0 100 369 203 314
1 100 228 784 366
b c d e
2 200 811 664 202
3 200 531 932 575
answered Nov 13 '18 at 16:34
user2921352user2921352
988
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
add a comment |
0
0
0
You can use the groupby method:
from pandas import DataFrame
columns = ["b", "c", "d", "e"]
data = [[100, 369, 203, 314], [100, 228, 784, 366], [200, 811, 664, 202], [200, 531, 932, 575]]
df = DataFrame(data=data,columns=columns)
def split_dfs(df, col):
return [group[1] for group in list(df.groupby(col))]
dfs = split_dfs(df, "b")
for df_group in dfs:
print(df_group)
b c d e
0 100 369 203 314
1 100 228 784 366
b c d e
2 200 811 664 202
3 200 531 932 575
answered Nov 13 '18 at 16:34
user2921352user2921352
988
You can use the groupby method:
from pandas import DataFrame
columns = ["b", "c", "d", "e"]
data = [[100, 369, 203, 314], [100, 228, 784, 366], [200, 811, 664, 202], [200, 531, 932, 575]]
df = DataFrame(data=data,columns=columns)
def split_dfs(df, col):
return [group[1] for group in list(df.groupby(col))]
dfs = split_dfs(df, "b")
for df_group in dfs:
print(df_group)
b c d e
0 100 369 203 314
1 100 228 784 366
b c d e
2 200 811 664 202
3 200 531 932 575
answered Nov 13 '18 at 16:34
user2921352user2921352
988
edited Nov 14 '18 at 9:08
answered Nov 13 '18 at 16:34
user2921352user2921352
988
answered Nov 13 '18 at 16:34
user2921352user2921352
988
answered Nov 13 '18 at 16:34
user2921352user2921352
988
988
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
add a comment |
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
Perfect! thank you.
– Andy Thomas
Nov 14 '18 at 8:56
add a comment |
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1
to clarify: did you want a new df for every single unique value of items in the b column?
– Capn Jack
Nov 13 '18 at 16:14
have you looked into groupby?
– Christian Sloper
Nov 13 '18 at 16:15
I don't really need a new df, if I can get the same result but in other datastructure it will be fine.
– Andy Thomas
Nov 13 '18 at 16:15
d=x: y for x, y in df.groupby('b')– Wen-Ben
Nov 13 '18 at 16:16
d = dict(tuple(df.groupby('b')))?– jezrael
Nov 13 '18 at 16:17