如何使用Python将列值转换为Martix

Ask is-关于将列值转换为矩阵格式的任何建议，如下所示：

您可以从numpy中选中add.outer

您可以从numpy中选中add.outer

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您可以仅使用numpy来执行此操作

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您可以仅使用numpy来执行此操作

非熊猫：

items = ['A', 'B','C','D','E']

# Make combinations
pairs = [f"{colA[i]}{colA[j]}" for i in range(len(items)) for j in range(len(items))]

# Find max character count per combo
max_sz = max(map(len, pairs))

# Set initial row to items list
output = [[" "] + items]

# Append additional rows based on starting character
for c in items:
    tmp = [c] + [p for p in pairs if str(p).startswith(c)]
    output.append(tmp)

# Format with specified padding from max_sz
final = ""
for ln in output:
    final += " ".join([f"{i:>{max_sz}}" for i in ln]) + "\n"

print(final)

输出：

    A  B  C  D  E
 A AA AB AC AD AE
 B BA BB BC BD BE
 C CA CB CC CD CE
 D DA DB DC DD DE
 E EA EB EC ED EE

非熊猫：

items = ['A', 'B','C','D','E']

# Make combinations
pairs = [f"{colA[i]}{colA[j]}" for i in range(len(items)) for j in range(len(items))]

# Find max character count per combo
max_sz = max(map(len, pairs))

# Set initial row to items list
output = [[" "] + items]

# Append additional rows based on starting character
for c in items:
    tmp = [c] + [p for p in pairs if str(p).startswith(c)]
    output.append(tmp)

# Format with specified padding from max_sz
final = ""
for ln in output:
    final += " ".join([f"{i:>{max_sz}}" for i in ln]) + "\n"

print(final)

输出：

    A  B  C  D  E
 A AA AB AC AD AE
 B BA BB BC BD BE
 C CA CB CC CD CE
 D DA DB DC DD DE
 E EA EB EC ED EE

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简单的解决方案我认为：

import numpy as np
m = np.repeat(list("ABCDE"), 5).reshape(5,5)
output = np.char.add(m, m.T)

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简单的解决方案我认为：

import numpy as np
m = np.repeat(list("ABCDE"), 5).reshape(5,5)
output = np.char.add(m, m.T)

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您的数据格式是什么？您的数据格式是什么？您没有定义df heredf可以定义为--lst=[A，B，C，D，E]df=pd.DataFramelst，columns=['col']您没有定义df heredf可以定义为--lst=[A，B，C，D，E]df=pd.DataFramelst，columns=['col']