Question

如何用文本中的字母替换所有等于1的值？

如果文本不完全适合数组，如何自动生成与第一个相同的新数组以瞄准包含所有字符？

提前致谢。

text = "helloworldddd"
my_array = [ [0,1,1],[1,1,0],[1,0,0] ]

result expecting: generated arrays which contains all characters
my_array2 = [ [0,h,e],[l,l,0],[o,0,0] ]
my_array3 = [ [0,w,o],[r,l,0],[d,0,0] ]
my_array4 = [ [0,d,d],[d,1,0],[1,0,0] ]

Answer 1

使用列表和复制，您应该阅读浅层和深层复制的区别，否则您只需复制列表引用 - 而不是值。在此处阅读更多内容：Deep copy a list in Python

计算你需要多少重复是很容易的，计算模式中的1的数量，除以字符串的长度，并添加1以获得良好的度量（整数除法是floor_division），如果它将与休息分开（用mod来检查）。

深入了解一个足够长的结果列表并开始用字符替换1：

from itertools import chain  # chaining multiple lists together
import copy                  # deepcopying the array to decouple copies

def createArr(t:str,arr:list):
    """Puts each character of t on the 1s of arr, extends arr with itself as pattern"""

    ones = sum(1 if q==1 else 0 for n in arr for q in n) # how many 1 in pattern
    mult = len(t)//ones  # how many reps needed for our text?
    if len(t)%ones != 0:
        mult += 1

    # create sufficient space by deepcopying the pattern into a longer list
    rv = list(chain(*(copy.deepcopy(arr) for _ in range(mult))))

    # start in sublinst on index 0
    subidx = 0
    # replace the 1s
    for c in t:    # walk all characters
        # maybe advance sublist index - if no 1s left
        while 1 not in rv[subidx] and subidx<len(rv):
            subidx+=1
        sub = rv[subidx] # get sublist
        sub[sub.index(1)] = c  # replace first 1 with act. char

    return rv 


text = "helloworldddd"
my_array = [ [0,1,1],[1,1,0],[1,0,0] ]


print(createArr(text,my_array))

输出：

 [[0, 'h', 'e'], ['l', 'l', 0], ['o', 0, 0], 
  [0, 'w', 'o'], ['r', 'l', 0], ['d', 0, 0], 
  [0, 'd', 'd'], ['d', 1, 0],   [1, 0, 0]]

你可以根据需要进行分区，将arr的长度分成相等的部分，这个排名很高的问题及其答案会告诉你如何：How do you split a list into evenly sized chunks?

如何用python中的文本替换2D数组中的值

1 个答案: