筛选 管道函数 如同r语言中的tidyverse.filter
df.query(" 列名 判断 ")
series.to_frame.query(" 列名 判断 ")生成元素重复的列表 如同r语言中的rep和rep_len函数
# R语言中 rep(x, times = 1, length.out = NA, each = 1)
# times 所有元素循环 Python实现
[1,2,3]*2
# 或者
# tile函数功能：对整个数组进行复制拼接 numpy.tile(a, reps)
arr_before = np.array([1,2,3])
np.tile(arr_before, 2)
# each 每个元素重复 Python实现
# repeat函数功能：对数组中的元素进行连续重复复制 numpy.repeat(a, repeats, axis=None)
np.repeat([1,2,], 2)
# 或者
list_before = [1, 2, 3]
list_after = [val for val in list_before for i in range(3)]https://www.qiniu.com/qfans/qnso-46166933
https://blog.csdn.net/sinat_26811377/article/details/98469964
https://blog.csdn.net/zyl1042635242/article/details/43052403
列表降维 sum()函数的妙用
newlist = sum(oldlist,[])Pandas 以表格样式显示 DataFrame
https://www.delftstack.com/zh/howto/python-pandas/pandas-display-dataframe-in-a-table-style/
np.tile(A, reps)
If reps has length
d, the result will have dimension ofmax(d, A.ndim).If
A.ndim < d, A is promoted to be d-dimensional by prepending new axes. So a shape (3,) array is promoted to (1, 3) for 2-D replication, or shape (1, 1, 3) for 3-D replication. If this is not the desired behavior, promote A to d-dimensions manually before calling this function.If
A.ndim > d, reps is promoted to A.ndim by pre-pending 1’s to it. Thus for an A of shape (2, 3, 4, 5), a reps of (2, 2) is treated as (1, 1, 2, 2).Note : Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.
Repeat elements of an array.
Broadcast an array to a new shape
https://numpy.org/doc/stable/reference/generated/numpy.tile.html
# 如果A的维数为1
# if A.ndim == 1
# 如果reps的长度为0
# reps.ndim < A.ndim
import numpy as np
np.tile( 0 , '') # array(0)
np.tile([0], '') # array([0])
np.tile([0], ()) # array([0])
# 如果reps的长度为1
# reps.ndim == A.ndim
# 此时开始 A = 0 或 A = [0] 输出结果一致
np.tile([0], 0) # array([], dtype=int32)
np.tile([0], 1) # array([0])
np.tile([0], 2) # array([0, 0])
# 如果reps的长度为2
# reps.ndim > A.ndim
# 可以发现3条规律:
# 1
# 由于np.array([0,0])和np.array((0,0))的结果都是array([0, 0])
# 所以reps的形式可以是[]也可以是() 效果一致
# 2
# np.tile([0], [0,0]) # array([], shape=(0, 0), dtype=int32)
# np.tile([0], [0,1]) # array([], shape=(0, 1), dtype=int32)
# 可见reps中只要有一个元素为0 返回结果为[] 但shape会随着reps改变
# 3
# []与() 维数为0
# 单个元素(包含空字符串'')与[单个元素]与(单个元素) 维数为1
np.tile([0], [1,2])
# array([[0, 0]])
np.tile([0], [2,1])
# array([[0],
# [0]])
np.tile([0], [2,2])
# array([[0, 0],
# [0, 0]])
# 如果A的维数... reps的长度...足与不足...