查找并计算列表中某个范围内数字的所有出现次数和位置
Find and count all occurrences and position of numbers in a range in a list
我想知道每个数字在 6 个数字集的列表中的每个索引位置出现的次数，但我不知道这些数字是什么，但它们的范围仅为 0-99。
示例列表：
data = [['22', '45', '6', '72', '1', '65'], ['2', '65', '67', '23', '98', '1'], ['13', '45', '98', '4', '12', '65']]
最终我会将结果计数放入 pandas DataFrame 中，看起来像这样：
num numofoccurances position numoftimesinposition
01 02 04 01
01 02 05 01
02 01 00 01
04 02 03 01
06 01 02 01
12 01 04 01
13 01 00 01
and so on...
由于每次出现在不同索引位置的 num 重复，结果数据会略有不同，但希望这有助于您理解我在寻找什么。
到目前为止，这是我开始的：
data = json.load(f)
numbers = []
contains = []
'''
This section is simply taking the data from the json file and putting it all into a list of lists containing the 6 elements I need in each list
'''
for i in data['data']:
item = [i[9], i[10]]
# print(item)
item = [words for segments in item for words in segments.split()]
numbers.append(item)
'''
This is my attempt to count to number of occurrences for each number in the range then add it to a list.
'''
x = range(1,99)
for i in numbers:
if x in i and not contains:
contains.append(x)
import pandas as pd
num_pos = [(num,pos) for i in data for pos,num in enumerate(i)]
df = pd.DataFrame(num_pos,columns = ['number','position']).assign(numoftimesinposition = 1)
df = df.astype(int).groupby(['number','position']).count().reset_index()
df1 = df.groupby('number').numoftimesinposition.sum().reset_index().\
rename(columns = {'numoftimesinposition':'numofoccurences'}).\
merge(df, on='number')
print(df1)
number numofoccurences position numoftimesinposition
0 1 2 4 1
1 1 2 5 1
4 2 1 0 1
7 4 1 3 1
9 6 1 2 1
2 12 1 4 1
3 13 1 0 1
5 22 1 0 1
6 23 1 3 1
8 45 2 1 2
10 65 3 1 1
11 65 3 5 2
12 67 1 2 1
13 72 1 3 1
14 98 2 2 1
15 98 2 4 1
如果上面的代码感觉很慢，那么使用 Counter from collections:
import pandas as pd
from collections import Counter
num_pos = [(int(num),pos) for i in data for pos,num in enumerate(i)]
count_data = [(num,pos,occurence) for (num,pos), occurence in Counter(num_pos).items()]
df = pd.DataFrame(count_data, columns = ['num','pos','occurence']).sort_values(by='num')
df['total_occurence'] = [Counter(df.num).get(num) for num in df.num]
print(df)
这应该可以解决您的查询（应该比极慢的 groupby（您将需要其中的 2 个）和其他 pandas 更大数据的操作更快）-
#get the list of lists into a 2d numpy array
dd = np.array(data).astype(int)
#get vocab of all unique numbers
vocab = np.unique(dd.flatten())
#loop thru vocab and get sum of occurances in each index position
df = pd.DataFrame([[i]+list(np.sum((dd==i).astype(int), axis=0)) for i in vocab])
#rename cols
df.columns = ['num', 0, 1, 2, 3, 4, 5]
#create total occurances of the item
df['numoccurances'] = df.iloc[:,1:].sum(axis=1)
#Stack the position counts and rename cols
stats = pd.DataFrame(df.set_index(['num','numoccurances']).\
stack()).reset_index().\
set_axis(['num', 'numoccurances', 'position', 'numtimesinposition'], axis=1)
#get only rows with occurances
stats = stats[stats['numtimesinposition']>0].reset_index(drop=True)
stats
num numoccurances position numtimesinposition
0 1 2 4 1
1 1 2 5 1
2 2 1 0 1
3 4 1 3 1
4 6 1 2 1
5 12 1 4 1
6 13 1 0 1
7 22 1 0 1
8 23 1 3 1
9 45 2 1 2
10 65 3 1 1
11 65 3 5 2
12 67 1 2 1
13 72 1 3 1
14 98 2 2 1
15 98 2 4 1
如结果所示-
1在您分享的示例数据中一共出现了2次，在第5位和第6位各出现了1次。同理2一共出现1次，也是第1位。
我想知道每个数字在 6 个数字集的列表中的每个索引位置出现的次数，但我不知道这些数字是什么，但它们的范围仅为 0-99。
示例列表：
data = [['22', '45', '6', '72', '1', '65'], ['2', '65', '67', '23', '98', '1'], ['13', '45', '98', '4', '12', '65']]
最终我会将结果计数放入 pandas DataFrame 中，看起来像这样：
num numofoccurances position numoftimesinposition
01 02 04 01
01 02 05 01
02 01 00 01
04 02 03 01
06 01 02 01
12 01 04 01
13 01 00 01
and so on...
由于每次出现在不同索引位置的 num 重复，结果数据会略有不同，但希望这有助于您理解我在寻找什么。
到目前为止，这是我开始的：
data = json.load(f)
numbers = []
contains = []
'''
This section is simply taking the data from the json file and putting it all into a list of lists containing the 6 elements I need in each list
'''
for i in data['data']:
item = [i[9], i[10]]
# print(item)
item = [words for segments in item for words in segments.split()]
numbers.append(item)
'''
This is my attempt to count to number of occurrences for each number in the range then add it to a list.
'''
x = range(1,99)
for i in numbers:
if x in i and not contains:
contains.append(x)
import pandas as pd
num_pos = [(num,pos) for i in data for pos,num in enumerate(i)]
df = pd.DataFrame(num_pos,columns = ['number','position']).assign(numoftimesinposition = 1)
df = df.astype(int).groupby(['number','position']).count().reset_index()
df1 = df.groupby('number').numoftimesinposition.sum().reset_index().\
rename(columns = {'numoftimesinposition':'numofoccurences'}).\
merge(df, on='number')
print(df1)
number numofoccurences position numoftimesinposition
0 1 2 4 1
1 1 2 5 1
4 2 1 0 1
7 4 1 3 1
9 6 1 2 1
2 12 1 4 1
3 13 1 0 1
5 22 1 0 1
6 23 1 3 1
8 45 2 1 2
10 65 3 1 1
11 65 3 5 2
12 67 1 2 1
13 72 1 3 1
14 98 2 2 1
15 98 2 4 1
如果上面的代码感觉很慢，那么使用 Counter from collections:
import pandas as pd
from collections import Counter
num_pos = [(int(num),pos) for i in data for pos,num in enumerate(i)]
count_data = [(num,pos,occurence) for (num,pos), occurence in Counter(num_pos).items()]
df = pd.DataFrame(count_data, columns = ['num','pos','occurence']).sort_values(by='num')
df['total_occurence'] = [Counter(df.num).get(num) for num in df.num]
print(df)
这应该可以解决您的查询（应该比极慢的 groupby（您将需要其中的 2 个）和其他 pandas 更大数据的操作更快）-
#get the list of lists into a 2d numpy array
dd = np.array(data).astype(int)
#get vocab of all unique numbers
vocab = np.unique(dd.flatten())
#loop thru vocab and get sum of occurances in each index position
df = pd.DataFrame([[i]+list(np.sum((dd==i).astype(int), axis=0)) for i in vocab])
#rename cols
df.columns = ['num', 0, 1, 2, 3, 4, 5]
#create total occurances of the item
df['numoccurances'] = df.iloc[:,1:].sum(axis=1)
#Stack the position counts and rename cols
stats = pd.DataFrame(df.set_index(['num','numoccurances']).\
stack()).reset_index().\
set_axis(['num', 'numoccurances', 'position', 'numtimesinposition'], axis=1)
#get only rows with occurances
stats = stats[stats['numtimesinposition']>0].reset_index(drop=True)
stats
num numoccurances position numtimesinposition
0 1 2 4 1
1 1 2 5 1
2 2 1 0 1
3 4 1 3 1
4 6 1 2 1
5 12 1 4 1
6 13 1 0 1
7 22 1 0 1
8 23 1 3 1
9 45 2 1 2
10 65 3 1 1
11 65 3 5 2
12 67 1 2 1
13 72 1 3 1
14 98 2 2 1
15 98 2 4 1
如结果所示-
1在您分享的示例数据中一共出现了2次，在第5位和第6位各出现了1次。同理2一共出现1次，也是第1位。